Sample Exercises

Hint

Given,
\(
\mathrm{h}=100 \mathrm{~m}, \mathrm{v}=20 \mathrm{~m} / \mathrm{s}, \mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^{2}
\)
This shows projectile,
a) time it takes to reach the ground,
\(
\begin{aligned}
&\mathrm{t}=\sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}} \\
&=\sqrt{\frac{2 \times 100}{9.8}}=4.51 \mathrm{~s}
\end{aligned}
\)
b) The horizontal distance traveled before reaching the ground,
\(
\begin{aligned}
&x=u t \\
&=20 \times 4.5=90 \mathrm{~m}
\end{aligned}
\)

c) Horizontal velocity remains constant throughout the frequency,
At “A”, \(\mathrm{V}_{\mathrm{x}}=20 \mathrm{~m} / \mathrm{s}, \mathrm{V}_{\mathrm{y}}=\mathrm{u}+\) at \(=0+9.8 \times 4.5=44.1 \mathrm{~m} / \mathrm{s}\)
Final velocity
\(
=\mathrm{v}_{\mathrm{r}}=\sqrt{(4.41)^{2}+(20)^{2}}=48.42 \mathrm{~m} / \mathrm{s}
\)
\(
\tan \beta=\frac{\mathrm{V}_{\mathrm{y}}}{\sqrt{\mathrm{x}}}=\frac{4.41}{20}=2.205
\)
\(
\Rightarrow \beta=\tan ^{-1} 2.205=66^{\circ}
\)

Hint

Given \(\mathrm{u}=40 \mathrm{~m} / \mathrm{s}, \theta=60^{\circ}\) and \(g=10 \mathrm{~m} / \mathrm{s}^{2}\)
Maximum Height \(\mathrm{h}_{\mathrm{m}}=\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{~g}}=60 \mathrm{~m}\)
Range \(\mathrm{R}=\frac{\mathrm{u}^{2} \sin 2 \theta}{\mathrm{g}}=80 \sqrt[2]{3} \mathrm{~m}\)

Hint

The acceleration due to gravity on earth is \(g=9.8 \mathrm{~m} / \mathrm{s}^{2} = 32.2 \mathrm{~ft} / \mathrm{s}^{2}\)
We know that 40 yards \(=120\) feet
Horizontal range \(\mathrm{x}=120 \mathrm{ft}\)
The initial speed of the ball is \(\mathrm{u}=64 \mathrm{ft} / \mathrm{s}\)
The angle at which the ball is kicked \(\theta=45^{\circ}\)
know that horizontal range \(x=u \cos \theta t\)
\(t=\frac{x}{u \cos \theta}\)
\(t=\frac{120}{64 \cos 45^{0}}\)
\(t=\frac{120}{64 \times \frac{1}{\sqrt{2}}}\)
\(
=\frac{120 \times 1.414}{64}
\)
\(
=2.65 \mathrm{sec}
\)

\(
y=u \sin \theta(t)-1 / 2 g{t}^{2}=64 \frac{1}{\sqrt{2}(2.65)}-\frac{1}{2}(32.2)(2.65)^{2}
\)
\(=7.08 \mathrm{ft}\) which is less than the height of goal post.
In time \(2.65\), the ball travels horizontal distance \(120 \mathrm{ft}(40 \mathrm{yd})\) and vertical height \(7.08 \mathrm{ft}\) which is less than \(10 \mathrm{ft}\). The ball will reach the goal post.

Hint

Goli moves like a projectile.
\(
\mathrm{h}=19.6 \mathrm{~cm}=0.196 \mathrm{~m}
\)
Horizontal distance \(\mathrm{x}=2 \mathrm{~m}\)
Time to reach the ground \(=\mathrm{t}=\sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}}=\frac{2 \times 0.196}{9.8}=0.2 \mathrm{sec}\)
Let the horizontal velocity with which the goli is to be projected be u
\(
\mathrm{u}=\mathrm{x} / \mathrm{t}=\frac{2 \mathrm{~m}}{0.2 \mathrm{~s}}=10 \mathrm{~m} / \mathrm{s}
\)

Hint

Width of the ditch \(=11.7 \mathrm{ft}\)
Length of the bike \(=5 \mathrm{ft}\)
The approach road makes an angle of \(15^{\circ}(\theta)\) with the horizontal.
Total horizontal range that should be covered by the biker to cross the ditch safely, \(R=11.7+5=16.7 \mathrm{ft}\)
Acceleration due to gravity, \(a=g=9.8 \mathrm{~m} / \mathrm{s}=32.2 \mathrm{ft} / \mathrm{s}^{2}\)
We know that the horizontal range is given by
\(R=\frac{u^{2} \sin 2 \theta}{g}\)
By putting respective values, we get:
\(
\begin{aligned}
&u^{2}=\frac{R g}{\sin 2 \theta}=\frac{16.7 \times 32.2}{\sin 30^{\circ}} \\
&\Rightarrow u \approx 32 f t / s
\end{aligned}
\)
Therefore, the minimum speed with which the motorbike should be moving is \(32 \mathrm{ft} / \mathrm{s}\).

Hint

Given:
Height (h) of the cliff \(=171 \mathrm{ft}\)
Horizontal distance from the bottom of the cliff \(=228 \mathrm{ft}\)
As per the question, the person throws the packet directly aiming to his friend at the initial speed (u) of \(15.0 \mathrm{ft} / \mathrm{s}\).

From the diagram, we can write:
\(
\begin{aligned}
&\tan \theta=\frac{171}{228} \\
&\Rightarrow \theta=\tan ^{-1}\left(\frac{171}{228}\right) \\
&\therefore \theta=37^{\circ}
\end{aligned}
\)
When the person throws the packet from the top of the cliff, it moves in projectile motion.
\(a=g=32.2 \mathrm{ft} / \mathrm{s}^{2}\) (Acceleration due to gravity)
Using the second equation of motion, we get:
\(
\begin{aligned}
&y=u \sin \theta t+\frac{1}{2} g t^{2} \\
&y=171 f t \\
&\theta=37^{\circ}
\end{aligned}
\)

\(
\begin{aligned}
&g=32 f t / s^{2} \\
&T=\text { Time of flight } \\
&171=15 \sin (37) t+\frac{1}{2} \times 32 \times t^{2}
\end{aligned}
\)
On solving this quadratic equation in \(\mathrm{t}\), we get:
\(
\begin{aligned}
&t=2.99 \mathrm{~s} \\
&\text { Range }=15 \cos (37) \times 2.99=35.81 \mathrm{ft}
\end{aligned}
\)
Distance by which the packet will fall short \(=228-35.81=192  \mathrm{ft}\)(approx.)

Hint

Here,
\(
\begin{aligned}
\mathrm{u} &=15 \mathrm{~m} / \mathrm{s} \\
\theta &=60^{\circ} \\
\mathrm{g} &=9.8 \mathrm{~m} / \mathrm{s}^{2}
\end{aligned}
\)
Formula for range of the ball
\(
\mathrm{x}=\frac{\mathrm{u}^{2} \sin 2 \theta}{\mathrm{g}}
\)
\(
=\frac{15^{2} \sin 2 \times 60}{9.8}
\)
\(
=\frac{225 \sin 120}{9.8}
\)
\(
\therefore \mathrm{x}=19.8 \mathrm{~m}
\)
So in case-1, the wall is \(5 \mathrm{~m}\) away from the projection point so it is in the horizontal range of the projectile. So the ball will hit the wall.

In case-2, \(22 \mathrm{~m}\) away the wall is not within the horizontal range. So the ball would not hit the wall.

Hint

Given:
The initial velocity of the projectile \(=u\)
The angle of projection \(=\theta\)
To find: Average velocity of the projectile
Average velocity
\(=\frac{\text { Change in displacement }}{\text { Time }}\)
Consider the projectile motion in the figure given below.
By the symmetry of the figure, it can be said that the line joining points \(A\) and \(B\) is horizontal.
So, there will be no effect on the vertical component of velocity of the projectile during displacement \(A B\).
We know that the projectile moves at a constant velocity \(u \cos \theta\) in the horizontal direction.
Hence, the average velocity of the projectile is \(u \cos \theta\).

Hint

Given:
Acceleration of the car \(=1 \mathrm{~m} / \mathrm{s}^{2}\)
Projection velocity of the ball (considered as a projectile) in the vertical direction \(=9.8 \mathrm{~m} / \mathrm{s}\)
Angle of projection, \(\theta=90^{\circ}\)
Let \(u\) be the initial velocity of the car when the ball is thrown.
Both the car and the ball have the same horizontal velocity.

Initial velocity of car is = Horizontal velocity of ball.
Distance travelled by ball \( S_{b}=u t\) (in horizontal direction)
and also the distanced travelled by the car in horizontal direction \(S_{c}=u t+1 / 2\) at \({ }^{2}\) where \(t \rightarrow\) time of flight of ball in air.
\(\therefore\) Car has travelled extra distance \(\mathrm{S}_{\mathrm{c}}-\mathrm{S}_{\mathrm{b}}=1 / 2 \mathrm{at}^{2}\).
Ball can be considered as a projectile having \(\theta=90^{\circ}\).
\(\therefore t=\frac{2 \mathrm{u} \sin \theta}{\mathrm{g}}=\frac{2 \times 9.8}{9.8}=2 \mathrm{sec}\).
\(
\therefore \mathrm{S}_{\mathrm{c}}-\mathrm{S}_{\mathrm{b}}=1 / 2 \mathrm{at}^{2}=2 \mathrm{~m}
\)
\(\therefore\) The ball will drop \(2 \mathrm{~m}\) behind the boy.

Hint

To directly hit the lowest plane, the ball should just touch point E.
Let point A be the origin of reference coordinate. Let \(u\) be the minimum speed of the ball.
We have:
\(
\begin{aligned}
&x=40 \mathrm{~cm} \\
&y=-20 \mathrm{~cm} \\
&\theta=0^{\circ} \\
&g=10 \mathrm{~m} / \mathrm{s}^{2}=1000 \mathrm{~cm} / \mathrm{s}^{2} \\
&\therefore y=x \tan \theta-g \frac{x^{2} \sec ^{2} \theta}{2 u^{2}} \\
&\Rightarrow-20=-\frac{800000}{2 u^{2}} \\
&\Rightarrow u=200 \mathrm{~cm} / \mathrm{s}=2 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
Thus, the minimum horizontal velocity of the ball is \(2 \mathrm{~m} / \mathrm{s}\).

Hint

Velocity of truck \(\mathrm{v}_{\mathrm{t}}=14.7 \mathrm{~m} / \mathrm{s}\)
Distance \(\mathrm{s}=58.8 \mathrm{~m}\)
(a) As seen from the truck the ball moves vertically upward and comes back. Time taken is equal to the time taken by truck to cover \(58.8 \mathrm{~m}\). Therefore angle of projection as seen from the truck \(=90^{\circ}\)

Time taken by the truck,  \(\text {t}=\frac{\mathrm{s}}{\mathrm{v_{t}}}=\frac{58.8}{14.7}=4 \mathrm{~s}\)
\(
\begin{aligned}
&\mathrm{t}=\frac{2 \mathrm{u} \sin \theta}{\mathrm{g}} \\
&\mathrm{u}=\frac{\operatorname{tg}}{2 \sin \theta}=\frac{4 \times 9.8}{2 \times 1}=2 \times 9.8=19.6 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)

(b) b) From the road it seems to be projectile motion.
Total time of flight \(=4 \mathrm{sec}\)
In this time horizontal range covered \(58.8 \mathrm{~m}=\mathrm{x}\)
\(
\begin{aligned}
&\therefore \mathrm{x}=\mathrm{u} \cos \theta \mathrm{t} \\
&\Rightarrow \mathrm{u} \cos \theta=14.7
\end{aligned}
\)
Taking the vertical component of velocity into consideration.
\(\begin{aligned} y=\frac{0^{2}-(19.6)^{2}}{2 \times(-9.8)}=19.6 \mathrm{~m}[\text { from }(\mathrm{a})] \\ \therefore \mathrm{y}=\mathrm{u} \sin \theta \mathrm{t}-1 / 2 \mathrm{gt}^{2} \\ \Rightarrow 19.6=\mathrm{u} \sin \theta(2)-1 / 2(9.8) 2^{2} \Rightarrow 2 \mathrm{u} \sin \theta=19.6 \times 2 \\ \Rightarrow \mathrm{u} \sin \theta=19.6 \\ \frac{\mathrm{u} \sin \theta}{\mathrm{u} \cos \theta}=\tan \theta \Rightarrow \frac{19.6}{14.7}=1.333 \\ \Rightarrow \theta=\tan ^{-1}(1.333)=53^{\circ} \\ \text { Again } u \cos \theta=14.7 \\ \Rightarrow u=\frac{14.7}{u \cos 53^{\circ}}=24.42 \mathrm{~m} / \mathrm{s} . \end{aligned}\)
The speed of ball is \(42.42 \mathrm{~m} / \mathrm{s}\) at an angle \(53^{\circ}\) with horizontal as seen from the road.

Hint

Given:
The angle of projection of the ball, \(\theta=53^{\circ}\)
Width and height of the bench \(=1 \mathrm{~m}\)
Initial speed of the ball \(=35 \mathrm{~m} / \mathrm{s}\)
Distance of the first bench from the batsman \(=110 \mathrm{~m}\)
The batsman strikes the ball \(1 \mathrm{~m}\) above the ground.
Let the ball land on the \(\mathrm{n}^{\text {th }}\) bench.
\(
\therefore y=(n-1) \quad \ldots(\mathrm{i})
\)
and,
\(
x=110+n-1=110+y
\)

Now ,
\(y=x \tan \theta-\left(\frac{g x^{2} \sec ^{2} \alpha}{2 u^{2}}\right)\) \(\Rightarrow y=(110+y)\left(\frac{4}{3}\right)-\frac{10 \times(110+y)^{2}\left(\sec ^{2} 53^{\circ}\right)}{2 \times(35)^{2}}\) \(=\frac{440}{3}+\frac{4}{3} y-\frac{250(110+y)^{2}}{18 \times(35)^{2}}\) \(=\frac{440}{3}+\frac{4}{3} y-\frac{250(110+y)^{2}}{18 \times 35^{2}}\)

Solving the above equation, we get: \(y=5\) \(\Rightarrow\) n -1=5 \(\Rightarrow\) n \(=6\)
The ball will hit the sixth bench of the gallery.

Hint

Length of the boat \(=1.0 \mathrm{~m}\)
Distance between the man and the centre of the boat \((R)=5.5 \mathrm{~m}\)
Initial speed (u) of throwing the apple by the man \(=10 \mathrm{~m} / \mathrm{s}\)
Acceleration due to gravity \((\mathrm{g})=10 \mathrm{~m} / \mathrm{s}^{2}\)
We know that the horizontal range is given by
\(\begin{aligned}
&R=\frac{u^{2} \sin 2 \theta}{g} \\
&\Rightarrow 5=\frac{(10)^{2} \sin 2 \theta}{10} \\
&\Rightarrow \sin 2 \theta=\frac{1}{2} \\
&\Rightarrow \theta=15^{\circ} \text { or } 75^{\circ}
\end{aligned}\)

Similarly, for the endpoint of the boat, we have:
Horizontal range \((R)=6 \mathrm{~m}\)
\(
\begin{aligned}
&R=\frac{u^{2} \sin 2 \theta}{g} \\
&\Rightarrow 6=\frac{(10)^{2} \sin 2 \theta}{10} \\
&\Rightarrow \sin 2 \theta=\frac{3}{5} \\
&\Rightarrow \theta=18^{\circ} \text { or } 71^{\circ}
\end{aligned}
\)
For a successful shot, the angle of projection \(\theta\) with initial speed \(10 \mathrm{~m} / \mathrm{s}\) may vary from \(15^{\circ}\) to \(18^{\circ}\) or from \(71^{\circ}\) to \(75^{\circ}\). The minimum angle is \(15^{\circ}\) and the maximum angle is \(75^{\circ}\).

Hint

a) Here the boat moves with the resultant velocity \(V_{r}\). But the vertical component \(10 \mathrm{~m} / \mathrm{s}\) takes him to the opposite shore.
\(\operatorname{tan} \theta=2 / 10=1 / 5\)
Velocity \(=10 \mathrm{~m} / \mathrm{s}\)
distance \(=400 \mathrm{~m}\)
Time \(=400 / 10=40 \mathrm{sec}\).
b) The boat will reach at point \(C\).
\(
\begin{aligned}
& \text { In } \triangle A B C, \tan \theta=\frac{B C}{A B}=\frac{B C}{400}=\frac{1}{5} \\
\Rightarrow & B C=400 / 5=80 \mathrm{~m} .
\end{aligned}
\)

Hint

(a) Width of the river \(=500 \mathrm{~m}\)
Rate of flow of the river \(=5 \mathrm{~km} / \mathrm{h}\)
Swimmer’s speed with respect to water \(=3 \mathrm{~km} / \mathrm{h}\)
As per the question, the swimmer heads in a direction making an angle \(\theta\) with the flow.
We know that the vertical component of velocity \(3 \sin \theta\) takes him to the opposite side of the river.
Distance to be travelled \(=0.5 \mathrm{~km}\)
Vertical component of velocity \(=3 \sin \theta \mathrm{km} / \mathrm{h}\)
Thus, we have:
\(
\text { Time }=\frac{\text { Distance }}{\text { Velocity }}=\frac{0.5}{3 \sin \theta} h=\frac{500 \times 6}{5 \sin \theta}=\frac{600}{\sin \theta} s
\)
\(\therefore\) Required time \(=\frac{10}{\sin \theta}\) minute.

(b) Here vertical component of velocity i.e. \(3 \mathrm{~km} / \mathrm{hr}\) takes him to opposite side Time \(=\frac{\text { Distance }}{\text { Velocity }}=\frac{0.5}{3}=0.16 \mathrm{hr}\)
\(5 \mathrm{~km} / \mathrm{h}\)
\(
\therefore 0.16 \mathrm{hr}=60 \times 0.16=9.6=10 \text { minute. }
\)

Hint

Distance between points A and B=\(500 \mathrm{~km}\)
B from A is \(30^{\circ}\) east of north.
Speed of wind due north, \(V_{w}=20 \mathrm{~m} / \mathrm{s}\)
Velocity of aeroplane is, \(V_{a}=150 \mathrm{~m} / \mathrm{s}\)
Let \(\vec{R}\) be the resultant direction of the plane to reach point B.

In \(\triangle A C D\) according to the sine formula
\(
\begin{aligned}
&\therefore \frac{20}{\sin A}=\frac{150}{\sin 30^{\circ}} \Rightarrow \sin A=\frac{20}{150} \sin 30^{\circ}=\frac{1}{15} \\
&\Rightarrow A=\sin ^{-1}(1 / 15)
\end{aligned}
\)
a) The direction is \(\sin ^{-1}(1 / 15)\) east of the line AB.
b) \(\sin ^{-1}(1 / 15)=3^{\circ} 48^{\prime}\)
\(
\begin{aligned}
&\Rightarrow 30^{\circ}+3^{\circ} 48^{\prime}=33^{\circ} 48^{\prime} \\
&R=\sqrt{150^{2}+20^{2}+2(150) 20 \cos 33^{\circ} 48^{\prime}}=167 \mathrm{~m} / \mathrm{s} . \\
&\text { Time }=\frac{\mathrm{s}}{\mathrm{v}}=\frac{500000}{167}=2994 \mathrm{sec}=50 \mathrm{~min} (approx.).
\end{aligned}
\)

Hint

The velocity of sound \(v\), Velocity of air \(u\), Distance between A and B be \(x\).
In the first case, the resultant velocity of sound \(=v+u\)
\(
\begin{aligned}
&\Rightarrow(v+u) t_{1}=x \\
&\Rightarrow v+u=x / t_{1} \quad \text {…(1) }
\end{aligned}
\)
In the second case, resultant velocity of sound \(=v-u\)

\(\begin{aligned}
&\therefore(v-u) t_{2}=x \\
&\Rightarrow v-u=x / t_{2}\quad \text {…(2) }
\end{aligned}\)
From (1) and (2) \( 2 v=\frac{x}{t_{1}}+\frac{x}{t_{2}}=x\left(\frac{1}{t_{1}}+\frac{1}{t_{2}}\right)\)
\(
\Rightarrow v=\frac{x}{2}\left(\frac{1}{t_{1}}+\frac{1}{t_{2}}\right)
\)
From (i) \(u=\frac{x}{t_{1}}-v=\frac{x}{t_{1}}-\left(\frac{x}{2 t_{1}}+\frac{x}{2 t_{2}}\right)=\frac{x}{2}\left(\frac{1}{t_{1}}-\frac{1}{t_{2}}\right)\)
\(\therefore\) Velocity of air \(v=\frac{x}{2}\left(\frac{1}{t_{1}}+\frac{1}{t_{2}}\right)\)
And velocity of wind \(u=\frac{x}{2}\left(\frac{1}{t_{1}}-\frac{1}{t_{2}}\right)\)

Hint

As per the question, each particle maintains a direction towards the particle at the next corner. So, particles will meet at centroid \(O\) of triangle ABC. Now, at any instant, the particles will form an equilateral triangle \(B A C\) with the same centroid \(O\). We know that \(B\) approaches \(A\), \(A\) approaches \(C\) and so on.
Now, we will consider the motion of particle B. Its velocity makes an angle of \(60^{\circ}\).
This component is the rate of decrease of distance BO.
Relative velocity between \(\mathrm{B}\) and \(\mathrm{A}\):

\(
\mathrm{v}_{\mathrm{r}}=\mathrm{v}-\mathrm{v} \cos 60^{\circ}=\mathrm{v}-\frac{\mathrm{v}}{2}=\frac{\mathrm{v}}{2}\)
Relative velocity = Total displacement/Time taken
\(
\begin{aligned}
&\Rightarrow \frac{\mathrm{v}}{2}=\frac{\mathrm{a}}{\mathrm{t}} \\
&\Rightarrow \mathrm{t}=\frac{2 \mathrm{a}}{\mathrm{v}}
\end{aligned}
\)
Hence, the time taken by the particles to meet each other is \(\frac{2 \mathrm{a}}{\mathrm{v}}\)

Hint

Displacement is given by the minimum distance between the initial and final positions of a particle. In the given case, all the girls start from point \(P\) and reach point \(Q\). The magnitudes of their displacements will be equal to the diameter of the ground.
Radius of the ground \(=200 \mathrm{~m}\)
Diameter of the ground \(=2 \times 200=400 \mathrm{~m}\)
Hence, the magnitude of the displacement for each girl is \(400 \mathrm{~m}\). This is equal to the actual length of the path skated by girl B.

Hint

(a) Displacement is given by the minimum distance between the initial and final positions of a body. In the given case, the cyclist comes to the starting point after cycling for 10 minutes. Hence, his net displacement is zero.
(b) Average velocity is given by the relation:
Average velocity \(=\frac{\text { Net displacement }}{\text { Total time }}\)
Since the net displacement of the cyclist is zero, his average velocity will also be zero.

(c) Average speed of the cyclist is given by the relation:
Average speed \(=\frac{\text { Total path length }}{\text { Total time }}\)
Total path length \(=O P+P Q+Q O\)
\(
\begin{aligned}
&=1+\frac{1}{4}(2 \pi \times 1)+1 \\
&=2+\frac{1}{2} \pi=3.570 \mathrm{~km}
\end{aligned}
\)
Time taken \(=10 \mathrm{~min}=\frac{10}{60}=\frac{1}{6} h\)
\(\therefore\) Average speed \(=\frac{3.570}{\frac{1}{6}}=21.42 \mathrm{~km} / \mathrm{h}\)

Hint

The path followed by the motorist is a regular hexagon with side \(500 \mathrm{~m}\), as shown in the given figure. Let the motorist start from point A. The motorist takes the third turn at D. Magnitude of displacement \(=\mathrm{AD}=\mathrm{BC}+\mathrm{EF}=500+500=1000 \mathrm{~m}\). Total path length \(=\mathrm{AB}+\mathrm{BC}+\mathrm{CD}=500+500+500=1500 \mathrm{~m}\). The motorist takes the sixth turn at point A, which is the starting point. Magnitude of displacement =0. Total path length \(=6 \mathrm{AB}\)
\( =6 \times 500=3000 \mathrm{~m}\)
The motorist takes the eight turn at point C
The magnitude of displacement \(=\mathrm{AC}\)
From triangle law, the magnitude of displacement is \(866.03 \mathrm{~m}\) at an angle of \(30^{\circ}\) with \(\mathrm{AC}\). It means AC makes an angle of \(30^{\circ}\) with the initial direction. Total path length = \(8 \times 500=4000 \mathrm{~m}\).

Hint

Here,
\(v_{c}=\) Velocity of the cyclist
\(v_{r}=\) Velocity of falling rain
In order to protect herself from the rain, the woman must hold her umbrella in the direction of the relative velocity \((v)\) of the rain with respect to the woman.
\(\vec{v}=\overrightarrow{v_{r}}+\left(-\vec{v}_{c}\right)\)
Resultant, \(|v|=\sqrt{v_{r^{2}}+v_{c^{2}}}\)
\(|v|=\sqrt{30^{2}+10^{2}}\)
\(|v|=\sqrt{900+100}\)
\(|v|=\sqrt{1000}=10 \sqrt{10} m\)
\(\tan \theta=\frac{v_{c}}{v_{r}}=\frac{10}{30}\)
\(\theta=\tan ^{-1}\left(\frac{1}{3}\right)\)
\(\theta=\tan ^{-1}(0.333) \approx 18^{\circ}\)

Hence, the woman must hold the umbrella toward the south, at an angle of nearly \(18^{\circ}\)

Hint

Speed of the man, \(v_{\mathrm{m}}=4 \mathrm{~km} / \mathrm{h}\)
Width of the river \(=1 \mathrm{~km}\)
Time taken to cross the river \(=\frac{\text { Width of river }}{\text { Speed of man }}\) \(=\frac{1}{4} h=\frac{1}{4} \times 60=15 \mathrm{~min}\)
Speed of the river, \(v_{r}=3 \mathrm{~km} / \mathrm{h}\)
Distance covered with flow of the river \(=v_{r} \times t\)
\(
\begin{aligned}
&=3 \times \frac{1}{4}=\frac{3}{4} k m \\
&=\frac{3}{4} \times 1000=750 m
\end{aligned}
\)

Hint

Velocity of the boat, \(v_{\mathrm{b}}=51 \mathrm{~km} / \mathrm{h}\)
Velocity of the wind, \(v_{\mathrm{w}}=72 \mathrm{~km} / \mathrm{h}\)
The flag is fluttering in the north-east direction. It shows that the wind is blowing toward the north-east direction. When the ship begins sailing toward the north, the flag will move along the direction of the relative velocity \(\left(v_{w b}\right)\) of the wind with respect to the boat. The angle between \(v_{\mathrm{w}}\) and \(\left(-v_{\mathrm{b}}\right)=90^{\circ}+45^{\circ}\)
\(
\begin{aligned}
&\tan \beta=\frac{51 \sin (90+45)}{72+51 \cos (90+45)} \\
&=\frac{51 \sin 45}{72+51(-\cos 45)}=\frac{51 \times \frac{1}{\sqrt{2}}}{72-51 \times \frac{1}{\sqrt{2}}} \\
&=\frac{51}{72 \sqrt{2}-51}=\frac{51}{72 \times 1.414-51}=\frac{51}{50.800} \\
&\therefore \beta=\tan ^{-1}(1.0038)=45.11^{\circ}
\end{aligned}
\)
Angle with respect to the east direction \(=45.11^{\circ}-45^{\circ}=0.11^{\circ}\).
Hence, the flag will flutter almost due east.

Hint

Height of the ceiling \(=25 \mathrm{~m}=\) Maximum height attained by the ball.
Velocity of the ball, \(\mathrm{u}=40 \mathrm{~m} / \mathrm{s}\)
Maximum height is given by \(\mathrm{h_{m}}=\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 g}\)
\(
\begin{aligned}
&\text { i.e., } \sin ^{2} \theta=\frac{2 \mathrm{gH}}{\mathrm{u}^{2}}=\frac{2 \times 9.8 \times 25}{(40)^{2}}=0.30625 \\
&\therefore \quad \sin \theta=\sqrt{0.30625}=0.5534 \\
&\Rightarrow \quad \theta=33.6
\end{aligned}
\)
\(\therefore\) Maximum horizontal distance,
\(
\begin{aligned}
R &=\frac{u^{2} \sin 2 \theta}{g}=\frac{(40)^{2} \sin 67.2}{9.8} \\
&=150.5 \mathrm{~m}
\end{aligned}
\)

Hint

Maximum horizontal distance, \(\mathrm{R_{m}}=100 \mathrm{~m}\)
The cricketer will only be able to throw the ball to the maximum horizontal distance when the angle of projection is \(45^{\circ}\), i.e. \(\theta=45^{\circ}\).

The max horizontal range for a projection velocity \(v\) is given by the relation:
\(
\mathrm{R}_{m }=\frac{\mathrm{u}^{2}}{\mathrm{~g}}
\)
\(
100=\frac{\mathrm{u}^{2}}{\mathrm{~g}}
\)
The ball will achieve the maximum height when it is thrown vertically upward. For such motion, the final velocity \(v\) is zero at the maximum height \(\mathrm{h_{m}}\). Acceleration, \(\mathrm{a}=\mathrm{-g}\)
Using the third equation of motion:
\(
\mathrm{v}^{2}-\mathrm{u}^{2}=-2 \mathrm{gh_{m}}
\)
\(
\mathrm{h_{m}}=\mathrm{u}^{2} / 2 \mathrm{~g}=100 / 2=50 \mathrm{~m}
\)

Hint

Here \(r=80 \mathrm{~cm}=0.8 \mathrm{~m}\);
\(
\begin{aligned}
&v=14 / 25 \mathrm{rev} / \mathrm{s} \\
&\omega=2 \pi v=2 \times \frac{22}{7} \times \frac{14}{25} \mathrm{rad} / \mathrm{s}=\frac{88}{25} \mathrm{rad} \mathrm{s}^{-1}
\end{aligned}
\)
The centripetal acceleration
\(
a=\omega^{2} r=\left(\frac{88}{25}\right)^{2} \times 0.80=9.90 \mathrm{~ms}^{-2}
\)
The direction of centripetal acceleration is along the string directed towards the centre of circular path.

Hint

Radius of the loop, \(r=1 \mathrm{~km}=1000 \mathrm{~m}\)
Speed of the aircraft,  \(v=900 \mathrm{~km} / \mathrm{h}=900 \mathrm{x} 5 / 18=250 \mathrm{~m} / \mathrm{s}\)
Centripetal acceleration, \(a_{c}=\frac{v^{2}}{r}\)
\(
=\frac{(250)^{2}}{1000}=62.5 \frac{m}{s^{2}}
\)
Acceleration due to gravity, \(\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^{2}\)
\(
\begin{aligned}
\frac{a_{c}}{g} &=\frac{62.5}{9.8}=6.38 \\
a_{c} &=6.38 g
\end{aligned}
\)

Hint

(a) Here
\(
\vec{r}(t)=\left(3.0 t \hat{i}-2.0 t^{2} \hat{j}+4.0 \hat{k}\right) \mathrm{m}
\)
\(\therefore\)
\(
\vec{v}(t)=\frac{\overrightarrow{d r}}{d t}=(3.0 \hat{i}-4.0 t \hat{j}) \mathrm{m} / \mathrm{s}
\)
and
\(
\vec{a}(t)=\frac{\overrightarrow{d v}}{d t}=(-4.0 \hat{j}) \mathrm{m} / \mathrm{s}^{2}
\)

(b) Magnitude of velocity at \(t=2.0 \mathrm{~s}\),
\(
\begin{aligned}
v_{(t=2 s)} &=\sqrt{(3.0)^{2}+(-4.0 \times 2)^{2}}=\sqrt{9+64}=\sqrt{73} \\
&=8.54 \mathrm{~m} \mathrm{~s}^{-1}
\end{aligned}
\)
This velocity will subtend an angle \(\beta\) from \(x\)-axis, where \(\tan \beta=\frac{(-4.0 \times 2)}{(3.0)}=-2.667\). \(=-2.6667 .\)
\(\therefore \quad \beta=\tan ^{-1}(-2.6667)=-69.44^{\circ}=69.44^{\circ}\) from negative \(x\)-axis.

Hint

The initial velocity of particle is \(\overrightarrow{\mathrm{u}}=10 \mathrm{jm} / \mathrm{s}\)
The acceleration of the particle is \(\vec{a}=8 \hat{i}+2 \hat{j}\)
(a) The position of the particle at any instant is given by:
\(
\overrightarrow{\mathrm{s}}=\overrightarrow{\mathrm{u} t}+\frac{1}{2} \overrightarrow{\mathrm{a}} \mathrm{t}^{2}
\)
\(
x \hat{i}+y \hat{j}=4 t^{2} \hat{i}+\left(10 t+t^{2}\right) \hat{j}
\)
Comparing the \(\mathrm{x}\) and \(\mathrm{y}\) component of the position:
\(x=4 t^{2}\)
\(
x=16 \Rightarrow t=2 s
\)
\(
y=10 t+t^{2}
\)
At \(\mathrm{t}=2, \mathrm{y}=24 \mathrm{~m}\)

(b) The speed of the particle at any instant is given by:
\(
\begin{aligned}
&\vec{v}=\vec{u}+\vec{a} t \\
&\vec{v}=10 \hat{j}+(8 \hat{i}+2 \hat{j}) 2
\end{aligned}
\)
\( \vec{v}=14 \hat{j}+16 \hat{i}\)
\(
|\vec{v}|=\sqrt{14^{2}+16^{2}}=21.26 \mathrm{~m} / \mathrm{s}
\)

Hint

In the Figure below “O” is the observation point at the ground. A and B are the positions of aircraft for which \(\angle \mathrm{AOB}=30^{\circ}\). Draw a perpendicular \(O C\) on \(A B\). Here \(O C=3400 \mathrm{~m}\) and \(\angle A O C\) \(=\angle \mathrm{COB}=15^{\circ}\).
In \(\triangle \mathrm{AOC}, A C=O C \tan 15^{\circ}=3400 \times 0.2679=910.86 \mathrm{~m}\). \(A B=A C+C B=A C+A C=2 A C=2 \times 910.86 \mathrm{~m}\)
Speed of the aircraft, \(v=\frac{\text { distance } A B}{\text { time }}=\frac{2 \times 910.86}{10}= 182.17 \mathrm{~ms}^{-1}=182.2 \mathrm{~ms}^{-1}\)

Hint

Here \(R=3 \mathrm{~km}=3000 \mathrm{~m}, \theta=30^{\circ}, g=9.8 \mathrm{~m} \mathrm{~s}^{-2}\).
As \(R=\frac{u^{2} \sin 2 \theta}{g}\)
\(\begin{aligned} \Rightarrow & 3000 =\frac{u^{2} \sin 2 \times 30^{\circ}}{9.8}=\frac{u^{2} \sin 60}{9.8} \\ \Rightarrow & u^{2} =\frac{3000 \times 9.8}{\sqrt{3} / 2}=3464 \times 9.8 \\ \text { Also, } & R^{\prime} =\frac{u^{2} \sin 2 \theta^{\prime}}{g} \Rightarrow 5000=\frac{3464 \times 9.8 \times \sin 2 \theta}{9.8} \end{aligned}\)
\(
\text { i.e., } \quad \sin 2 \theta^{\prime}=\frac{5000}{3464}=1.44
\)
which is impossible because sine of an angle cannot be more than 1 . Thus this target cannot be hoped to be hit.

Hint

Velocity of plane, \(v_{p}=720 \times 5 / 180 \mathrm{~ms}^{-1}=200 \mathrm{~ms}^{-1}\)
Velocity of shell \(=600 \mathrm{~ms}^{-1}\);
\(
\begin{aligned}
\sin \theta &=\frac{200}{600}=\frac{1}{3} \\
\theta &=\sin ^{-1}\left(\frac{1}{3}\right)=19.47^{\circ}
\end{aligned}
\)
This angle is with the vertical. Let \(h_{min}\) be the required minimum height.
Using equation
\(
\begin{aligned}
v^{2}-u^{2} =2 \text { as, we get } \\
(0)^{2}-(600 \cos \theta)^{2} =-2 \times 10 \times h_{min} \\
h_{min} \quad =\frac{600 \times 600\left(1-\sin ^{2} \theta\right)}{20} \\
=30 \times 600\left(1-\frac{1}{9}\right)=\frac{8}{9} \times 30 \times 600 \mathrm{~m}=16 \mathrm{~km} .
\end{aligned}
\)

Hint

Here \(v=27 \mathrm{~km} / \mathrm{h}=27 \times \frac{5}{18} \mathrm{~m} / \mathrm{s}=7.5 \mathrm{~m} / \mathrm{s}, r=80 \mathrm{~m}\) and tangential acceleration \(a_{t}=-0.50 \mathrm{~m} / \mathrm{s}^{2}\)
\(\therefore\) Centripetal acceleration \(a_{c}=\frac{v^{2}}{r}=\frac{(7.5)^{2}}{80}=0.70 \mathrm{~ms}^{-2}\) (radially inwards).
Thus, as shown in fig. above, two accelerations are acting in mutually perpendicular directions. If \(\vec{a}\) be the resultant acceleration, then
\(
|\vec{a}|=\sqrt{a_{t}^{2}+a_{c}^{2}}=\sqrt{(0.5)^{2}+(0.7)^{2}}=0.86 \mathrm{~ms}^{-2}
\)

and \(\quad \tan \beta=\frac{a_{c}}{a_{t}}=\frac{0.7}{0.5}=1.4\)
\(\Rightarrow \quad \beta=\tan ^{-1}(1.4)=54.5^{\circ}\) from the direction of negative of the velocity.

Hint

(a) Let the projectile be fired at an angle \(\theta\) with \(x\)-axis.
As \(\theta\) depends on \(t, \theta(t)\), at any instant
\(
\begin{aligned}
\tan \theta(t)=\frac{v_{y}}{v_{x}}=\frac{v_{o y}-g t}{v_{o x}} \\ \quad\left(\text { Since } v_{y}=v_{o y}-g t \text { and } v_{x}=v_{o x}\right) \\
\theta(t) =\tan ^{-1}\left(\frac{v_{o y}-g t}{v_{o x}}\right)
\end{aligned}
\)

Since,
\(
h_{\max }=\frac{u^{2} \sin ^{2} \theta}{2 g}
\)

\(\begin{aligned} & \text { and } R =\frac{u^{2} \sin 2 \theta}{g} \\ \Rightarrow & \frac{h_{\max }}{R} =\frac{u^{2} \sin ^{2} \theta / 2 g}{u^{2} \sin 2 \theta / g}=\frac{\tan \theta}{4} \\ \Rightarrow & \frac{\tan \theta}{4} =\frac{h_{\max }}{R} \\ & \text { or } \tan \theta =\frac{4 h_{\max }}{R} \\ & \text { or } \theta =\tan ^{-1}\left(\frac{4 h_{\max }}{R}\right) . \end{aligned}\)

Hint

\(
\vec{a}=4 \vec{i}+3 \vec{j}, \vec{b}=3 \vec{i}+4 \vec{j}
\)
a) \(|\vec{a}|=\sqrt{4^2+3^2}=5\)
b) \(|\vec{b}|=\sqrt{9+16}=5\)
c) \(|\vec{a}+\overrightarrow{\mathrm{b}}|=|7 \overrightarrow{\mathrm{i}}+7 \overrightarrow{\mathrm{j}}|=7 \sqrt{2}\)
d)
\(
\begin{aligned}
& \vec{a}-\vec{b}=(-3+4) \hat{i}+(-4+3) \hat{j}=\hat{i}-\hat{j} \\
& |\vec{a}-\vec{b}|=\sqrt{1^2+(-1)^2}=\sqrt{2}
\end{aligned}
\)

Hint

The displacement of the car is represented by \(\overrightarrow{A D}\).
\(
\begin{aligned}
& \overrightarrow{A D}=2 \hat{i}+0.5 \hat{j}+4 \hat{i} \\
& =6 \hat{i}+0.5 \hat{j}
\end{aligned}
\)
Magnitude of \(\overrightarrow{A D}\) is given by
\(
\begin{aligned}
& A D=\sqrt{A E^2+D E^2} \\
& =\sqrt{6^2+(0.5)^2} \\
& =\sqrt{36+0.25}=6.02 \mathrm{~km}
\end{aligned}
\)

Now,
\(
\begin{aligned}
& \tan \theta=\frac{D E}{A E}=\frac{1}{12} \\
& \Rightarrow \theta=\tan ^{-1}\left(\frac{1}{12}\right)
\end{aligned}
\)
Hence, the displacement of the car is \(6.02 \mathrm{~km}\) along the direction \(\tan ^{-1}\left(\frac{1}{12}\right)\) with positive the \(x\)-axis.

Hint

\(\vec{a}\) is a vector of magnitude 4.5 unit due north.
a) \(3|\vec{a}|=3 \times 4.5=13.5\)
\(3 \vec{a}\) is along north having magnitude 13.5 units.
b) \(-4|\vec{a}|=-4 \times 4.5=-18\) unit
\(-4 \vec{a}\) is a vector of magnitude 18 unit due south.

Hint

We have:
\(
\begin{aligned}
& \vec{a}=2 \vec{i}+3 \vec{j}+4 \vec{k} \\
& \vec{b}=3 \vec{i}+4 \vec{j}+5 \vec{k}
\end{aligned}
\)
Using scalar product, we can find the angle between vectors \(\vec{a}\) and \(\vec{b}\). i.e.,
\(\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta\)
So, \(\theta=\cos ^{-1}\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}\right)\)
\(=\cos ^{-1}\left(\frac{2 \times 3+3 \times 4+4 \times 5}{\sqrt{\left(2^2+3^2+4^2\right)} \sqrt{\left(3^2+4^2+5^2\right)}}\right)\)
\(=\cos ^{-1}\left(\frac{38}{\sqrt{29} \sqrt{50}}=\cos ^{-1} \frac{38}{\sqrt{1450}}\right)\)
\(\therefore\) The required angle is \(\cos ^{-1} \frac{38}{\sqrt{1450}}\).

Hint

Given:
\(
\begin{aligned}
& \vec{A}=2 \hat{i}+3 \hat{j}+4 \hat{k} \text { and } \\
& \vec{B}=4 \hat{i}+3 \hat{j}+2 \hat{k}
\end{aligned}
\)
The vector product of \(\vec{A} \times \vec{B}\) can be obtained as follows:
\(
\begin{aligned}
& \vec{A} \times \vec{B}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & 3 & 4 \\
4 & 3 & 2
\end{array}\right| \\
& =\hat{i}(6-12)-\hat{j}(4-16)+\hat{k}(6-12) \\
& =-6 \hat{i}+12 \hat{j}-6 \hat{k}
\end{aligned}
\)

Hint

\(
y=3 x^2+6 x+7
\)
\(\therefore\) Area bounded by the curve, \(x\) axis with coordinates with \(x=5\) and \(x=10\) is given by,
\(
\text { Area } \left.\left.\left.=\int_0^y d y=\int_5^{10}\left(3 x^2+6 x+7\right) d x=3 \frac{x^3}{3}\right]_5^{10}+5 \frac{x^2}{3}\right]_5^{10}+7 x\right]_5^{10}=1135 \text { sq.units. }
\)

Hint

\(
\text { Area }=\int_0^y d y=\int_0^\pi \sin x d x=-[\cos x]_0^\pi=2
\)

Hint

The given function is \(y=e^{-x}\).
When \(x=0, y=e^{-0}=1\)
When \(x\) increases, the value of \(y\) decrease. Also, only when \(x=\infty, y=0\)
So, the required area can be determined by integrating the function from 0 to \(\infty\).
\(
\begin{aligned}
& \therefore \text { Area }=\int_0^{\infty} e^{-x} d x \\
& =-\left[e^{-x}\right]_0^{\infty} \\
& =-[0-1]=1 \mathrm{~sq} \cdot \mathrm{unit}
\end{aligned}
\)

Hint

According to the question, we have:
\(
\frac{d p}{d t}=(10 N)+(2 N / s) t
\)
Momentum is zero at time, \(\mathrm{t}=0\)
Now, \(d p=\left[(10 \mathrm{~N})+\left(2 \mathrm{Ns}^{-1}\right) t\right] d t\)
On integrating the above equation, we get:
\(
\begin{aligned}
& p=\int_0^{10} d p=\int_0^{10} 10 d t+\int_0^{10}(2 t d t) \\
& \Rightarrow p=\left[10 t+2 \frac{t^2}{2}\right]_0^{10} \\
& \Rightarrow p=10 \times 10+100-0 \\
& \therefore p=100+100=200 \mathrm{~kg} \mathrm{~m} / \mathrm{s}
\end{aligned}
\)

Hint

Changes in a function of \(y\) and the independent variable \(x\) are related as follows:
\(
\begin{aligned}
& \frac{d y}{d x}=x^2 \\
& \Rightarrow d y=x^2 d x
\end{aligned}
\)
Integrating of both sides, we get:
\(
\begin{aligned}
& \int d y=\int x^2 d x \\
& \Rightarrow y=\frac{x^3}{3}+c
\end{aligned}
\)
where \(c\) is a constant
\(\therefore y\) as a function of \(x\) is represented by
\(
y=\frac{x^3}{3}+c
\)

Hint

(a) 1001
Number of significant digits =4
(b) 100.1
Number of significant digits =4
(c) 100.10
Number of significant digits =5
(d) 0.001001
Number of significant digits =4

Hint

The metre scale is graduated at every millimetre.
i.e., \(1 \mathrm{~m}=1000 \mathrm{~mm}\)
The minimum number of significant digits may be one (e.g., for measurements like \(4 \mathrm{~mm}\) and \(6 \mathrm{~mm}\) ) and the maximum number of significant digits may be 4 (e.g., for measurements like \(1000 \mathrm{~mm}\) ). Hence, the number of significant digits may be \(1,2,3\) or 4 .

Hint

Length of the cylinder, \(l=4.54 \mathrm{~cm}\)
Radius of the cylinder, \(r=1.75 \mathrm{~cm}\)
Volume of the cylinder, \(V=\pi r^2 l=(\pi) \times(4.54) \times(1.75)^2\)

The minimum number of significant digits in a particular term is three. Therefore, the result should have three significant digits, while the other digits should be rounded off.
\(
\begin{aligned}
& \therefore \text { Volume, } V=\pi r^2 l \\
& =(3.14) \times(1.75) \times(1.75) \times(4.54) \\
& =43.6577 \mathrm{~cm}^3
\end{aligned}
\)
Since the volume is to be rounded off to 3 significant digits, we have:
\(
V=43.7 \mathrm{~cm}^3
\)

Hint

We know that,
Average thickness \(=\frac{2.17+2.17+2.18}{3}=2.1733 \mathrm{~mm}\)
Rounding off to 3 significant digits, average thickness \(=2.17 \mathrm{~mm}\).

Hint

As shown in the figure,
Actual effective length \(=(90.0+2.13) \mathrm{cm}\)
But, in the measurement \(90.0 \mathrm{~cm}\), the no. of significant digits is only 2 .
So, the addition must be done by considering only 2 significant digits of each measurement.
So, effective length \(=90.0+2.1=92.1 \mathrm{~cm}\).