Motion in a Plane (Projectile Motion-Entrance Paper) Quiz-L2

Hint

(b) The motion of the object shot in two cases can be depicted as below

Using third equation of motion, \(v^2=u^2-2 g h\)
As the object stops finally, so \(v=0\)
For inclined motion, \(g=g \sin \theta\) and \(h=x\)
Substituting these values in Eq. (i), we get
\(
\begin{array}{ll}
\Rightarrow & u^2=2 g \sin \theta x \\
\Rightarrow & x=\frac{u^2}{2 g \sin \theta}
\end{array}
\)
\(
\text { For case I: } \quad x_1=\frac{u^2}{2 g \sin 60^{\circ}}
\)
\(
\text { For case II: } \quad \begin{aligned}
x_2 & =\frac{u^2}{2 g \sin 30^{\circ}} \\
\because \quad \frac{x_1}{x_2} & =\frac{u^2}{2 g \sin 60^{\circ}} \times \frac{2 g \sin 30^{\circ}}{u^2} \\
& =\frac{1}{2} \times \frac{2}{\sqrt{3}}=\frac{1}{\sqrt{3}} \text { or } 1: \sqrt{3}
\end{aligned}
\)

Hint

(a) Given, distance between the two buildings, \(d=100 \mathrm{~m}\)
Height of each tower, \(h=200 \mathrm{~m}\)
Speed of each bullet, \(v=25 \mathrm{~ms}^{-1}\)
The situation can be shown as below.

where, \(x\) be the vertical distance travelled from the top of the building and \(t\) be the time at which they collide.
As two bullets are fired toward each other, so their relative velocity will be
\(
v_{\text {rel }}=25-(-25)=50 \mathrm{~ms}^{-1}
\)
Then, time, \(\quad t=\frac{d}{v_{\text {rel }}}=\frac{100}{50}=2 \mathrm{~s}\)
The distance or height at which they collide is calculated from equation of motion,
\(
x=u t+\frac{1}{2} a t^2
\)
The bullet is initially at rest, i.e. \(u=0\) and as it is moving under the effect of gravity \(a=-g\), so
The bullet is initially at rest, i.e. \(u=0\) and as it is moving under the effect of gravity \(a=-g\), so
\(
\begin{array}{l}
x=-\frac{1}{2} g t^2 \\
x=-\frac{1}{2} \times 10(2)^2=-20 \mathrm{~m}
\end{array}
\)
The negative sign shows that the bullets will collide \(20 \mathrm{~m}\) below the top of tower, i.e. at a height of \((200-20)=180 \mathrm{~m}\) from the ground after \(2 \mathrm{~s}\).

Hint

(c) To obtain maximum range, angle of projection must be \(45^{\circ}\), i.e. \(\theta=45^{\circ}\).
\(
\begin{aligned}
& R_{\max }=\frac{u^2 \sin \left(2 \times 45^{\circ}\right)}{g}=\frac{u^2}{g} \\
\because & H_{\max }=\frac{u^2 \sin ^2 45^{\circ}}{2 g}=\frac{u^2}{4 g}=\frac{R_{\max }}{4}
\end{aligned}
\)
So, \(H_{\max }\) is \(25 \%\) of \(R_{\max }\)

Hint

(a) Given, \(u=98 \mathrm{~ms}^{-1}\) and \(\theta=30^{\circ}\)
\(\because\) Range of a projectile,
\(
\begin{array}{l}
R=\frac{u^2 \sin (2 \theta)}{g}=\frac{98 \times 98 \times \sin 60^{\circ}}{9.8} \\
R=490 \sqrt{3} \mathrm{~m}
\end{array}
\)

Hint

(a) Let at any instant of time, the length \(A B\) be \(l\), here angle \(\theta\) and length \(l\) vary with time, then using Pythagoras theorem in \(\triangle A B C\),

\(
x^2+y^2=l^2
\)

On differentiating both sides w.r.t \(t\), we get or \(2 x \frac{d x}{d t}+2 y \frac{d y}{d t}=2 l \frac{d l}{d t}\)
As, there is no vertical motion of the block, so
\(
\frac{d y}{d t}=0, \frac{d x}{d t}=v_x \text { and } \frac{d l}{d t}=v
\)
\(
\begin{aligned}
2 x v_x & =2 l v \\
v_x & =\frac{l}{x} v \quad \text { or } \quad v_x=\frac{v}{\left(\frac{x}{l}\right)}=\frac{v}{\sin \theta}
\end{aligned}
\)

Hint

\(
\text { (a) Horizontal range, } R=\frac{u^2 \sin 2 \theta}{g}
\)

When \(\theta=45^{\circ}\), maximum horizontal range,
\(
R_{\max }=\frac{u^2}{g} \sin 90^{\circ}=\frac{u^2}{g}
\)
When \(\theta=135^{\circ}\), maximum horizontal range,
\(
R=\frac{u^2}{g} \sin 270^{\circ}=\frac{-u^2}{g}
\)
Negative sign implies opposite direction.

Hint

(d) Given, \(x=a \cos (p t), y=b \sin (p t)\) \(\therefore \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\), i.e. equation of ellipse
Now,
\(
\begin{array}{l}
\mathbf{r}=x \hat{\mathbf{i}}+y \hat{\mathbf{j}}=a \cos (p t) \hat{\mathbf{i}}+b \sin (p t) \hat{\mathbf{j}} \\
\mathbf{v}=\frac{d \mathbf{r}}{d t}=-p a \sin (p t) \hat{\mathbf{i}}+p b \cos (p t) \hat{\mathbf{j}} \\
\mathbf{a}=\frac{d \mathbf{v}}{d t}=-p^2 a \cos (p t) \hat{\mathbf{i}}-p^2 b \sin (p t) \hat{\mathbf{j}}
\end{array}
\)
\(
\text { At } \quad \begin{aligned}
t & =\frac{\pi}{2 p}, \\
\mathbf{v} & =-p a \hat{i} \text { and } \mathbf{a}=-p^2 \hat{\mathbf{j}}
\end{aligned}
\)
Thus, velocity is perpendicular to acceleration.
Also, \(a=-p^2 b\), i.e. directed towards a fixed point as \(p\) and \(b\) are positive constants.
\(
\begin{array}{ll}
\text { As, } & v=\frac{d r}{d t} \Rightarrow \Delta r=\int_0^{\frac{\pi}{2 p}} v d t \\
\Rightarrow & \Delta r=[a \cos (p t) \hat{\mathbf{i}}+b \sin (p t) \hat{\mathbf{j}}]_0^{\frac{\pi}{2 p}}=-a \hat{\mathbf{i}}+b \hat{\mathbf{j}} \\
\text { So, } & |\mathbf{r}|=\sqrt{a^2+b^2}
\end{array}
\)

Hint

(b) Position vector of the particle is given by
\(
\mathbf{r}=\cos \omega t \hat{\mathbf{x}}+\sin \omega t \hat{\mathbf{y}}
\)
where, \(\omega\) is a constant.
Velocity of the particle is
\(
\begin{aligned}
\mathbf{v} & =\frac{d \mathbf{r}}{d t}=\frac{d}{d t}(\cos \omega t \hat{\mathbf{x}}+\sin \omega t \hat{\mathbf{y}}) \\
& =(-\sin \omega t) \omega \hat{\mathbf{x}}+(\cos \omega t) \omega \hat{\mathbf{y}} \\
& =-\omega(\sin \omega t \hat{\mathbf{x}}-\cos \omega t \hat{\mathbf{y}})
\end{aligned}
\)
Acceleration of the particle,
\(
\begin{aligned}
\mathbf{a} & =\frac{d \mathbf{v}}{d t}=\frac{d}{d t}(-\omega \sin \omega t \hat{\mathbf{x}}+\omega \cos \omega t \hat{\mathbf{y}}) \\
& =-\omega^2 \cos \omega t \hat{\mathbf{x}}-\omega^2 \sin \omega t \hat{\mathbf{y}} \\
\Rightarrow \quad \mathbf{a} & =-\omega^2 \mathbf{r}=\omega^2(-\mathbf{r})
\end{aligned}
\)
Assuming the particle is at \(P\), then its position vector is directed as shown in the diagram.

Therefore, acceleration is directed towards \(-\mathbf{r}\), i.e. towards \(O\) (origin).
\(
\begin{aligned}
\mathbf{v} \cdot \mathbf{r} & =-\omega(\sin \omega t \hat{\mathbf{x}}-\cos \omega t \hat{\mathbf{y}}) \cdot(\cos \omega t \hat{\mathbf{x}}+\sin \omega t \hat{\mathbf{y}}) \\
& =-\omega[\sin \omega t \cdot \cos \omega t+0+0-\sin \omega t \cdot \cos \omega t] \\
& =-\omega(0)=0 \\
\Rightarrow \mathbf{v} & \perp \mathbf{r}
\end{aligned}
\)
Thus, velocity is perpendicular to \(\mathbf{r}\)

Hint

(a) The equation of trajectory,
\(
\begin{aligned}
y & =x \tan \alpha\left[1-\frac{x}{R}\right] \text { (Gives) } \\
Q & =P \tan \theta\left[1-\frac{P}{R}\right] \dots(i) \\
\text { and } \quad P & =Q \tan \theta\left[1-\frac{Q}{R}\right] \dots(ii)
\end{aligned}
\)

On dividing Eq. (i) by Eq. (ii), we get
\(
\begin{aligned}
\frac{Q^2}{P^2} & =\frac{[1-P / R]}{[1-Q / R]} \\
\frac{1}{R}\left[P^3-Q^3\right] & =P^2-Q^2
\end{aligned}
\)
\(
R=\frac{P^3-Q^3}{P^2-Q^2}=\frac{P^2+P Q+Q^2}{P+Q}
\)
\(
\begin{aligned}
\frac{Q}{P} & =\tan \theta\left[1-\frac{P(P+Q)}{P^2+P Q+Q^2}\right] \\
& =\tan \theta\left[\frac{P^2+P Q+Q^2-P^2-P Q}{P^2+P Q+Q^2}\right] \\
\tan \theta & =\frac{P^2+Q^2+P Q}{P Q} \\
\theta & =\tan ^{-1}\left[\frac{P^2+P Q+Q^2}{P Q}\right]
\end{aligned}
\)

Hint

(a) Consider the diagram below

For horizontal motion, \(50=10 \cos 30^{\circ} \times t\)
\(
\Rightarrow \quad t=\frac{5}{\cos 30^{\circ}}=\frac{5}{\frac{\sqrt{3}}{2}}=\frac{10}{\sqrt{3}}
\)
For vertical motion,
\(
\begin{aligned}
h & =10 \sin 30^{\circ} \times \frac{10}{\sqrt{3}}-\frac{1}{2} \times 10 \times\left(\frac{10}{\sqrt{3}}\right)^2 \\
& =10 \times \frac{1}{2} \times \frac{10}{\sqrt{3}}-\frac{100}{\sqrt{3}} \times \frac{5}{\sqrt{3}}=\frac{100}{\sqrt{3}}\left(\frac{1}{2}-\frac{5}{\sqrt{3}}\right) \\
& =\frac{50}{\sqrt{3}}\left[1-\frac{10}{\sqrt{3}}\right] \mathrm{m}
\end{aligned}
\)

Hint

(b)

\(
\begin{array}{l}
\text { Range }=\frac{\mathrm{u}^2 \sin 2 \theta}{\mathrm{g}} \\
\text { Range }=\frac{\mathrm{u}^2(2 \sin \theta \cos \theta)}{\mathrm{g}}=\frac{\mathrm{u}^2\left(2 \sin \left(90^{\circ}-\theta\right) \cos \left(90^{\circ}-\theta\right)\right)}{\mathrm{g}}
\end{array}
\)

Range of projectile is same for \(\theta\) and \(90^{\circ}-\theta\).

Given, \(\theta=40^{\circ}\)
Another angle \(=90^{\circ}-\theta=90^{\circ}-40^{\circ}=50^{\circ}\)

Hint

(a)

 Given, a particle having horizontal range is twice the greatest height attained by it, i.e. \(\mathrm{R}=2 \mathrm{H}\)
\(
\begin{array}{ll}
& \frac{v^2 \sin 2 \theta}{g}=\frac{2 v^2 \sin ^2 \theta}{2 g} \\
\Rightarrow & 2 \sin \theta \cos \theta=\sin ^2 \theta \Rightarrow \tan \theta=2 \\
\Rightarrow & \sin \theta=\frac{2}{\sqrt{5}} \Rightarrow \cos \theta=\frac{1}{\sqrt{5}} \\
\therefore \quad & R=\frac{v^2 \sin 2 \theta}{g}=\frac{v^2 \times 2 \sin \theta \cos \theta}{g} \\
\Rightarrow \quad & R=v^2 \times \frac{2}{g} \times \frac{2}{\sqrt{5}} \times \frac{1}{\sqrt{5}} \\
\text { i.e. } & R=\frac{4 v^2}{5 g}
\end{array}
\)

Hint

(d) Given, \(\theta_1=\theta_2=45^{\circ}\)
\(
\begin{array}{l}
\theta_1+\theta_2=\left(45^{\circ}+\alpha\right)+\left(45^{\circ}-\alpha\right)=90^{\circ} \\
u_1=u_2=u \Rightarrow R_1=R_2=\text { ? } \\
R_1=\frac{u_1^2 \sin \left(\theta_1+\theta_2\right)}{g} \\
\Rightarrow \quad R_1=\frac{u^2 \sin 90^{\circ}}{g}=\frac{u^2}{g} \dots(i)\\
\Rightarrow \quad R_2=\frac{u^2 \sin 90^{\circ}}{g}=\frac{u^2}{g} \\
\end{array}
\)

Similarly, \(R_2=\frac{u_2^2 \sin \left(\theta_1+\theta_2\right)}{g} \dots(ii)\)
\(
\Rightarrow \quad R_2=\frac{u^2 \sin 90^{\circ}}{g}=\frac{u^2}{g}
\)
The ratio of horizontal ranges
\(
\frac{R_1}{R_2}=\frac{u^2 / g}{u^2 / g}=1: 1 \Rightarrow \quad R_1: R_2=1: 1
\)

Hint

Average velocity,

\(
\begin{aligned}
\mathbf{v}_{\mathrm{av}} & =\frac{\text { Net displacement }}{\text { Time taken }} \\
& =\frac{(13-2) \hat{\mathbf{i}}+(14-3) \hat{\mathbf{j}}}{5} \\
& =\frac{11 \hat{\mathbf{i}}+11 \hat{\mathbf{j}}}{5}=\frac{11}{5}(\hat{\mathbf{i}}+\hat{\mathbf{j}})
\end{aligned}
\)

Hint

\(
\text { (c) For vertical motion, } h_1=u \sin \theta t_1-\frac{1}{2} g t_1^2 \text { (For } h_1 \text { ) }
\)
\(
\begin{array}{c}
t_1=\frac{h_1+\frac{1}{2} g t_1^2}{u \sin \theta} \dots(i) \\
h_2=u \sin \theta t_2-\frac{1}{2} g t_2^2 \text { (For } h_2 \text { ) }
\end{array}
\)
\(
t_2=\frac{h_2+\frac{1}{2} g t_2^2}{u \sin \theta} \dots(ii)
\)
On dividing Eq. (i) by Eq. (ii), we get
\(
\frac{t_1}{t_2}=\frac{\left(h_1+\frac{1}{2} g t_1^2\right) / u \sin \theta}{\left(h_2+\frac{1}{2} g t_2^2\right) / u \sin \theta}
\)
\(
\Rightarrow \quad h_1 t_2-h_2 t_1=\frac{g}{2}\left(t t_2^2-t_1^2 t_2\right)
\)
The time of flight of the ball,
\(
\begin{aligned}
T & =\frac{2 u \sin \theta}{g}=\frac{2}{g}(u \sin \theta) \\
& =\frac{2}{g}\left(\frac{h_1+1 / 2 g t_1^2}{t_1}\right) \text { [From Eq. (i)] }
\end{aligned}
\)
\(
\begin{array}{l}
=\frac{2}{t_1}\left(\frac{h_1}{g}+\frac{t_1^2}{2}\right) \\
=\frac{h_1}{t_1} \times \frac{2}{g}+t_1=\frac{h_1}{t_1} \times\left(\frac{t_1 t_2^2-t_1^2 t_2}{h_{t_2}-h_2 t_1}\right)+t_1
\end{array}
\)
\(
=\frac{h_1 t_1 t_2^2-h_1 t_1^2 t_2+h_1 t_1^2 t_2-h_2 t_1^3}{t_1\left(h_1 t_2-h_2 t_1\right)}=\left(\frac{h_1 t_2^2-h_2 t_1^2}{h_1 t_2-h_2 t_1}\right)
\)

Hint

(c) As we know that, maximum height of a projectile is given by
\(
H_{\max }=\frac{u^2 \sin ^2 \theta}{2 g}
\)
where, \(u=\) initial velocity of projectile
\(g=\) acceleration due to gravity
and \(\theta=\) angle of projection.
As per question,
\(
H_{\max }=\frac{v^2 \sin ^2 45^{\circ}}{2 g} \dots(i) \left(\text { As, } u=v, \theta=45^{\circ}\right)
\)
Now, range of a projectile is given by
\(
\begin{array}{l}
R=\frac{u^2 \sin 2 \theta}{g} \\
\Rightarrow \quad R=\frac{v^2 \sin \left(2 \times 45^{\circ}\right)}{g} \\
\Rightarrow \quad R=\frac{v^2 \sin 90^{\circ}}{g} \dots(ii) \\
\end{array}
\)
On dividing Eq. (i) by Eq. (ii), we get
\(
\begin{array}{l}
\frac{H_{\max }}{R}=\frac{v^2 \sin ^2 45^{\circ} \times g}{2 g \times v^2 \sin 90^{\circ}}=\frac{1}{4 \times 1} \\
\Rightarrow \quad R=4 H_{\max }=4 H \\
\end{array}
\)

Hint

(d) The horizontal component of velocity,
\(
v_x=u=10 \mathrm{~ms}^{-1}
[latex]

[latex]
\begin{array}{l}
\text { Now, } \tan \theta=\tan 45^{\circ}=\frac{v_y}{v_x}=\frac{v_y}{10} \\
\Rightarrow \quad v_y=10 \mathrm{~ms}^{-1}
\end{array}
\)

Hint

\(
\text { (b) As, } \quad t=\frac{2 u \sin \theta}{g} \Rightarrow 4=\frac{2 u \sin \theta}{g} \dots(i)
\)

\(
\begin{array}{l}
\Rightarrow \quad u \sin \theta=2 g \\
\text { and } \quad h=\frac{u^2 \sin ^2 \theta}{2 g}=\frac{4 g^2}{2 g} \quad \text { [From Eq.(i)] } \\
\Rightarrow \quad h=20 \mathrm{~m} \\
\end{array}
\)

Hint

(c) From the question’s figure, the \(x\)-component remain unchanged, while the \(y\)-component is reverse. Then, the velocity at point \(B\) is \((2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}) \mathrm{ms}^{-1}\).

Hint

(b) Time of flight, \(T=\frac{2 u \sin \theta}{g}\)

Given, \(u=u_0, T=1 \mathrm{~s}, \theta=30^{\circ}, g=9.8 \mathrm{~ms}^{-2}\)
\(
u_0=g=9.8 \mathrm{~ms}^{-1}
\)

Hint

(a) As the velocity makes an angle of \(30^{\circ}\) with horizontal, so the horizontal component of velocity at the instant will be \(v \cos 30^{\circ}\).
\(
\begin{array}{l}
\Rightarrow \quad v \cos 30^{\circ}=5 \\
\Rightarrow v=\frac{5}{\cos 30^{\circ}}=\frac{5}{\sqrt{3} / 2}=\frac{10}{\sqrt{3}} \mathrm{~ms}^{-1}
\end{array}
\)

Hint

(a) As the horizontal ranges are the same.
\(
\begin{array}{l}
\therefore \quad \frac{v_0^2 \sin 2 \theta_1}{g}=\frac{v_0^2 \sin 2 \theta_2}{g} \\
\text { So, } \quad \sin 2 \theta_1=\sin 2 \theta_2 \\
\text { or } \quad 2 \theta_1=\pi-2 \theta_2 \\
\Rightarrow \quad \theta_1+\theta_2=\pi / 2 \\
\therefore \quad\left(h_1\right)_{\max }=\frac{v_0^2 \sin ^2 \theta_1}{2 g} \\
\end{array}
\)
\(
\begin{array}{l}
\text { and } \quad\left(h_2\right)_{\max }=\frac{v_0^2 \sin ^2 \theta_2}{2 g} \\
\therefore \quad \frac{\left(h_1\right)_{\max }}{\left(h_2\right)_{\max }}=\frac{\sin ^2 \theta_1}{\sin ^2 \theta_2}=\frac{\sin ^2 \theta_1}{\cos ^2 \theta_1}=\tan ^2 \theta_1 \\
{\left[\because \sin ^2 \theta_2=\sin ^2\left(\frac{\pi}{2}-\theta_2\right)=\cos ^2 \theta / 2\right]}
\end{array}
\)

Hint

(a) Given, \(\mathbf{r}=2 t \hat{\mathbf{i}}+t^2 \hat{\mathbf{j}}\)

Velocity vector, \(\mathbf{v}=\frac{d \mathbf{r}}{d t}\) and using \(\frac{d}{d x} x^n=n x^{n-1}\)
We have, \(\frac{d \mathbf{r}}{d t}=2 \hat{\mathbf{i}}+2 t \hat{\mathbf{j}}\)

Hint

(b) Given, \(R=H\)
\(
\text { Range, } R=\frac{u^2(2 \sin \theta \cos \theta)}{g}
\)
Height, \(H=\frac{u^2 \sin ^2 \theta}{2 g}\)
Hence, \(\frac{u^2(2 \sin \theta \cos \theta)}{g}=\frac{u^2 \sin ^2 \theta}{2 g}\)
\(
\begin{aligned}
2 \cos \theta & =\frac{\sin \theta}{2} \\
\Rightarrow \quad \tan \theta & =4 \Rightarrow \theta=\tan ^{-1}(4)
\end{aligned}
\)

Hint

(d) Maximum height and time of flight depends upon the vertical component of initial velocity.
\(
\begin{array}{l}
H_1>H_2 \\
\Rightarrow \quad u_{y_1}>u_{y_2} \\
\text { Range } \quad R_2>R_1 \\
\text { So, } \quad u_2>u_1 \\
\end{array}
\)

Hint

(d) According to given condition,
\(
\begin{aligned}
\frac{h_1}{h_2} & =\frac{\sin ^2 \theta}{\sin ^2\left(90^{\circ}-\theta\right)}=\frac{20}{80} \\
\Rightarrow \quad \tan ^2 \theta & =\frac{1}{4} \Rightarrow \tan \theta=\frac{1}{2} \\
\sin \theta & =\frac{1}{\sqrt{5}} \text { and } \cos \theta=\frac{2}{\sqrt{5}}
\end{aligned}
\)
So, \(\quad h=\frac{u^2 \sin ^2 \theta}{2 g}\)
\(\Rightarrow \quad 20=\frac{u^2}{10 g}\) or \(\frac{u^2}{g}=200\)
\(\therefore \quad\) The range, \(R=\frac{u^2 \times 2 \sin \theta \cos \theta}{g}=\frac{200 \times 2 \times 2}{\sqrt{5} \times \sqrt{5}}\)
\(
=\frac{200 \times 4}{5}=160 \mathrm{~m}
\)

Hint

(b) Since, the ball reaches from one player to another in \(2 \mathrm{~s}\), so the time period of the flight, \(T=2 \mathrm{~s}\)
\(
\Rightarrow \quad \frac{2 u \sin \theta}{g}=2 \mathrm{~s}
\)

Here, \(u\) is the initial velocity and \(\theta\) is the angle of projection.
\(
\Rightarrow \quad u \sin \theta=g \dots(i)
\)

Now, we know that the maximum height of the projection,
\(
H=\frac{u^2 \sin ^2 \theta}{2 g} \text { or } H=\frac{(u \sin \theta)^2}{2 g}
\)

On putting the value of \(u \sin \theta\) from Eq. (i), we get
\(
H=\frac{g^2}{2 g}=\frac{g}{2} \text { or } H=\frac{g}{2}=\frac{10}{2} \mathrm{~m} \text { or } H=5 \mathrm{~m}
\)

Hint

(d) \(\left|\frac{d v}{d t}\right|=|a|=9.8 \mathrm{~ms}^{-2}=\) constant and it is true that in case of projectile motion, the magnitude of velocity first decreases and then increases during the motion.

Hint

(a) At highest point, velocity is horizontal and acceleration is vertical, i.e. both are perpendicular to each other and hence, their dot product is zero.

Hint

\(
\begin{array}{l}
\text { (b) } h_1=\frac{u^2}{2 g}, h_2=\frac{u^2 \sin ^2 30^{\circ}}{2 g}=\frac{u^2}{8 g} \\
\Rightarrow h_1=4 h_2
\end{array}
\)

Also, at highest point, \(v=0\) in first case and \(v \neq 0\) in second case.

Hint

(a) For both cases, \(t=\sqrt{\frac{2 h}{g}}=\) constant.

Because, initial vertical downward component of velocity will be zero for both the particles and both move under the effect of \(g\).

Hint

(a) Vertical component of velocity of ball at point \(P\).
\(
\begin{aligned}
u_V & =0+g t \\
& =10 \times 0.4=4 \mathrm{~ms}^{-1}
\end{aligned}
\)
Horizontal component of velocity \(=\) initial velocity
\(
\Rightarrow \quad v_H=4 \mathrm{~m} / \mathrm{s}
\)

So, the speed with which it hits the ground, \(v=\sqrt{v_H^2+v_V^2}=4 \sqrt{2} \mathrm{~m} / \mathrm{s}\) and \(\tan \theta=\frac{v_V}{v_H}=\frac{4}{4}=1 \Rightarrow \theta=45^{\circ}\)
It means the ball hits the ground at an angle of \(45^{\circ}\) to the horizontal.
Height of the table \(h=\frac{1}{2} g t^2=\frac{1}{2} \times 10 \times(0.4)^2=0.8 \mathrm{~m}\)
Horizontal distance travelled by the ball from the edge of table \(h=u t=4 \times 0.4=1.6 \mathrm{~m}\)

Hint

(c) The angle between instantaneous velocity vector and acceleration vector before attaining the maximum height is acute \((0\) to \(\pi / 2)\) and after is obtuse ( \(\pi / 2\) to \(\pi\) ). At the highest point, it is perpendicular \((\pi / 2)\).

Hint

(c)
\(
\begin{array}{l}
t_1=t_2=\sqrt{\frac{2 h}{g}} \quad \text { (where, } h=\text { height of tower) } \\
v_1=\sqrt{2 g h}
\end{array}
\)

While \(v_2=\sqrt{v_1^2+v_0^2}\) (where, \(v_0=\) initial horizontal velocity)
Therefore, both statements are correct.

Hint

(c)

A) average velocity between initial and final point is
\(
v_{\text {ave }}=\frac{R}{T}=\frac{u \cos \theta T}{T}=U \cos \theta
\)
B) change in velocity between initial and final point is
\(
g T=g \frac{2 u \sin \theta}{g}=2 u \sin \theta
\)
C) change in velocity between initial and peak point is
\(
g \frac{T}{2}=u \sin \theta
\)
D) average velocity between initial and peak point is
\(
\hat{i} u \cos \theta+\frac{\hat{j} u \sin \theta}{2}
\)

Hint

(d) In horizontal projectile motion,
Horizontal component of velocity,
\(
u_x=u=10 \mathrm{~ms}^{-1}
\)
Vertical component of velocity,
\(
u_y=g t=10 \times 1=10 \mathrm{~ms}^{-1}
\)
Horizontal displacement
\(
=u \times t=10 \times(1)=10 \mathrm{~m}
\)

Vertical displacement \(=\frac{1}{2} g t^2=\frac{1}{2} \times 10 \times(1)^2=5 \mathrm{~m}\)

Hint

\(
R=\frac{u^2 \sin 2 \theta}{g}
\)

For maximum range \(\theta=45^{\circ}, \sin 2 \theta=1\)
\(
\therefore R_{\max }=\frac{u^2}{g}
\)

Given, \(R=16 \mathrm{~km}=16 \times 10^3 \mathrm{~m}\)
\(
\begin{array}{l}
g=10 \mathrm{~m} / \mathrm{s}^2 \\
\Rightarrow u^2=16 \times 10^3 \times 10 \\
\Rightarrow u=400 \mathrm{~m} / \mathrm{s}
\end{array}
\)

Hint

A ball is projected at an angle of \(45^{\circ}\) with the horizontal. This is a projectile motion.
Let the initial velocity of the ball is \(u\).
Now, the horizontal component of the velocity will be, \(u \cos 45^{\circ}=\frac{u}{\sqrt{2}}\)
Again, the vertical component of the velocity will be, \(u \sin 45^{\circ}=\frac{u}{\sqrt{2}}\)
In a projectile motion, at the highest point of its flight, the vertical component of the velocity becomes zero and the velocity of the object is given only by the horizontal component of the velocity.
So, at the height of point of flight of a projectile motion, the velocity of the object will be \(\frac{u}{\sqrt{2}}\).
Now, initially the kinetic energy of the ball is \(E\) when the object starts moving with velocity \(u\).
So, we can write that,
\(
E=\frac{1}{2} m u^2
\)
Where, \(m\) is the mass of the ball.
Now, at the highest point of flight, the ball is moving with velocity \(\frac{u}{\sqrt{2}}\). so, its kinetic energy will be,
\(
E_f=\frac{1}{2} m\left(\frac{u}{\sqrt{2}}\right)^2=\frac{1}{4} m u^2
\)
Now, dividing the second equation with the first equation we get that,
\(
\begin{array}{l}
\frac{E_f}{E}=\frac{\frac{1}{4} m u^2}{\frac{1}{2} m u^2} \\
E_f=\frac{E}{2}
\end{array}
\)

So, the kinetic energy of the ball at its highest point of flight will be, \(\frac{E}{2}\).

Hint

Let the body be projected at an angle \(\theta\) with initial speed \(u\).
The maximum height attained by the body is given by
\(
H=\frac{u^2 \sin ^2 \theta}{2 g}
\)
And the range of the projectile,
\(
R=\frac{u^2 \sin 2 \theta}{g}=\frac{2 u^2 \sin \theta \cos \theta}{g}
\)
Now given \(R=3 H\)
\(
\begin{array}{l}
\Rightarrow \quad \frac{2 u^2 \sin \theta \cos \theta}{g}=\frac{3 \times u^2 \sin ^2 \theta}{2 g} \\
\Rightarrow \quad \frac{\sin \theta}{\cos \theta}=\tan \theta=\frac{4}{3} \\
\Rightarrow \theta=\tan ^{-1}\left(\frac{4}{3}\right)=53^{\circ}
\end{array}
\)

Hint

Range is same for angles of projection \(\theta\) and \((90-\theta)\)
\(
\begin{array}{l}
t_1=2 u \sin \theta / g \\
t_2=[2 u \sin (90-\theta)] / g=2 u \cos \theta / g .
\end{array}
\)
Hence, \(t_1 t_2=4\left(u^2 \sin \theta \cos \theta / g^2\right)=2 / g\left[u^2 \sin \theta / g\right]\)
\(=(2 / g) R\)., Where \({R}\) is range.
Hence, \(t_1 t_2 \propto R\) because \(R\) is constant.

Hint

As \(\theta_2=\left(90-\theta_1\right)\),
So range of projectile,
\(
\begin{array}{l}
R_1=\frac{v_0^2 \sin 2 \theta}{g}=\frac{v_0^2 2 \sin \theta \cos \theta}{g} \\
R_2=\frac{v_0^2 2 \sin \left(90-\theta_1\right) \cos \left(90-\theta_1\right)}{g} \\
R_2=\frac{v_0^2 2 \cos \theta_1 \sin \theta_1}{g}=R_1
\end{array}
\)

Hint

As per the problem we have to find the angle between the velocity and acceleration at the highest point. Now when a particle is projected an angle of \(\theta\), having some initial velocity u moving along a curved path under constant acceleration acting in downward direction can be represented as given,

As per the diagram \(\mathrm{H}\) is the highest point of the projectile motion and at the highest point the velocity acting is \(u \cos \theta\) and the acceleration acting on the particle is the constant acceleration and it is acting in downward direction and it the acceleration due to gravity. Hence acceleration and velocity are perpendicular to each other. Now from the above figure and from the above discussion we can conclude that the angle between the velocity and acceleration at the highest point is \(90^{\circ}\).

Hint

\(
\begin{array}{l}
\mathrm{F}=m a \quad \therefore a=\frac{{\mathrm{F}}}{m}=\frac{1}{3}\left(6 t^2 \hat{i}+4 t \hat{j}\right)=2 t^2 \hat{i}+\frac{4}{3} t \hat{j} \\
a=\frac{d \mathrm{~V}}{d t} \quad \therefore v=\int_0^t a d t=\int_0^3 2 t^2 \hat{i} d t+\int_0^3 \frac{4}{3} t \hat{j} d t=18 \hat{i}+6 \hat{j}
\end{array}
\)

Hint

Let \(u\) be the initial velocity of the projectile.
Range of a projectile is given by:
\(
\mathrm{R}=\frac{\mathrm{u}^2 \sin 2 \theta}{\mathrm{g}}
\)
Step 2: Ratio of \(R_1 \& R_2\)
At \(\theta_1=45^{\circ}-\theta, R=R_1\)
\(
\mathrm{R}_1=\frac{\mathrm{u}^2 \sin 2\left(45^{\circ}-\theta\right)}{\mathrm{g}}=\frac{\mathrm{u}^2 \sin \left(90^{\circ}-2 \theta\right)}{\mathrm{g}}
\)
At \(\theta_2=45^{\circ}+\theta, \mathrm{R}=\mathrm{R}_2\)
\(
R_2=\frac{u^2 \sin 2\left(45^{\circ}+\theta\right)}{g}=\frac{u^2 \sin \left(90^{\circ}+2 \theta\right)}{g}
\)
The ratio of \(R_1\) and \(R_2\) is given by
\(
\frac{R_1}{R_2}=\frac{\sin \left(90^{\circ}-2 \theta\right)}{\sin \left(90^{\circ}+2 \theta\right)}=\frac{\cos 2 \theta}{\cos 2 \theta}=1
\)
Hence the ratio of the range of the projectile is \(1: 1\).

Hint

Let, \(\theta\) be the angle which the particle makes with an \(\mathrm{x}\)-axis.

From the given figure:
\(
\tan \theta=\frac{y}{x}=\frac{3}{\sqrt{3}}=\sqrt{3} \Rightarrow \theta=\tan ^{-1} \sqrt{3}=60^{\circ} .
\)

Hint

The formula for momentum is given below:
\(
p=m v
\)
From the question, momentum change when the particle reaches ground level is asked.
Thus, the horizontal momentum remains unchanged.
\(
\begin{array}{l}
\Delta p=\Delta p_{\text {vertical }} \\
\Delta p_{\text {vertical }}=(\text { Final momentum })_{\text {vertical }}-(\text { Initial momentum })_{\text {vertical }} \\
\text { Initial momentum }=m v \sin 45 \\
\text { Final momentum }=-({mv} \sin 45) \\
\Delta p_{\text {vertical }}=-(m v \sin 45)-(m v \sin 45) \\
=-(2 m v \sin 45) \\
\Delta p_{\text {vertical }}=2 m v \times \frac{1}{\sqrt{2}} \\
\Delta p_{\text {vertical }}=\sqrt{2}(m v) .
\end{array}
\)

Hint

The horizontal range is the same when angle of projection is \(\theta\) or \(\left(90^{\circ}-\theta\right)\).

Hint

Let \(\mathrm{v}\) be velocity of a projectile at maximum height \(\mathrm{H}\)
\(
\mathrm{v}=\mathrm{u} \cos \theta
\)

According to given problem, \(\mathrm{v}=\frac{\mathrm{u}}{2}\)
\(
\begin{array}{l}
\therefore \frac{\mathrm{u}}{2}=\mathrm{u} \cos \theta \Rightarrow \cos \theta=\frac{1}{2} \\
\theta=60^{\circ}
\end{array}
\)

Hint

Step 1: Given data
Initial velocity, \(u=3 \hat{i}+4 \hat{j}\)
Acceleration, \(a=0.4 \hat{i}+0.3 \hat{j}\)
Time, \(t=10 \mathrm{~s}\)
Step 2: Find Final velocity
Final velocity \((v)\) is given by using the first equation of motion,
\(
\begin{aligned}
v & =u+a t \\
v & =3 \hat{i}+4 \hat{j}+(0.4 \hat{i}+0.3 \hat{j}) 10 \\
v & =7 \hat{i}+7 \hat{j} \\
V & =\sqrt{7^2+7^2} \\
V & =7 \sqrt{ } 2 \text { units }
\end{aligned}
\)

Hint

Given:
initial velocity \(\mathrm{u}=20 \mathrm{~m} / \mathrm{s}\)
\(
\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2
\)
We know that, for maximum range the angle of projection \(\theta=45^{\circ}\)
so,
\(
\mathrm{R}_{\max }=\frac{\mathrm{u}^2 \sin 2 \theta}{\mathrm{g}}=\frac{20^2 \sin 90^{\circ}}{10}=\frac{20 \times 20}{10}=40 \mathrm{~m} \text {. }
\)

Hint

Height of projectile
\(
H=\frac{u^2 \sin ^2 \theta}{2 g}=\frac{u^2 \sin ^2 45^{\circ}}{2 g}
\)
\(
H=\frac{u^2}{4 g} \ldots \ldots(i)
\)
Range of projectile \(R=\frac{u^2 \sin 2 \theta}{g}=\frac{u^2 \sin 90^{\circ}}{g}\) \(\Rightarrow R=\frac{u^2}{g}\)
\(
\begin{array}{l}
\tan \alpha=\frac{H}{R / 2} \\
\operatorname{In} \triangle O A B=\frac{u^2 / 4 g}{u^2 / 2 g} \\
\tan \alpha=\frac{1}{2} \Rightarrow \alpha=\tan ^{-1}\left(\frac{1}{2}\right)
\end{array}
\)

Hint

For shortest path(AB), velocity along river flow is zero. Horizontal compaonent should cancel out.
\(
\begin{array}{l}
20 \sin \theta=10 \Rightarrow \sin \theta=\frac{10}{20}=\frac{1}{2} \\
\theta=30^{\circ} \text { west }
\end{array}
\)

Hint

\(
\begin{array}{l}
\mathrm{H}_{\max }=\frac{\mathrm{u}^2 \sin ^2 \theta}{2 \mathrm{~g}} \\
=\frac{(280)^2\left(\sin 30^{\circ}\right)^2}{2(9.8)} \\
=1000 \mathrm{~m}
\end{array}
\)

Hint

At the top most point of its trajectory particle will have only horizontal component of velocity
\(
\begin{array}{l}
\mathrm{V}=v \cos \theta \\
=20 \times \frac{1}{2} \\
=10 \mathrm{~m} / \mathrm{s}
\end{array}
\)

Hint

Direction of velocity and centripetal acceleration changes continuously so \(\vec{v}\) is not constant and \(\vec{a}\) is not a constant.

Hint

Maximum height reached by the ball
\(
\begin{array}{l}
H=\frac{v^2 \sin ^2 \theta}{2 g} \\
=\frac{(20)^2 \sin ^2 30^{\circ}}{2 g} \\
H=5 m
\end{array}
\)

Hint

At highest point only horizontal component of velocity remains
\(
u_x=u \cos \theta=10 \cos 30^{\circ}=5 \sqrt{3} m s^{-1}
\)

Hint

\(
\begin{array}{l}
T=\frac{2 \pi R}{V} \\
V=\frac{2 \pi R}{T} \\
H=\frac{u^2 \sin ^2 \theta}{2 g} \\
4 R=\frac{4 \pi^2 R^2 \sin ^2 \theta}{T^2 2 g} \\
\sin ^2 \theta=\frac{8 R T^2 g}{4 \pi^2 R^2} \\
\sin ^2 \theta=\sqrt{\frac{2 T^2 g}{\pi^2 R}} \\
\theta=\sin ^{-1}\left(\frac{2 T^2 g}{\pi^2 R}\right)^{1 / 2}
\end{array}
\)

Hint

velocity of car at \(t=4 \mathrm{sec}\) is
\(
\begin{array}{l}
v=u+a t \\
v=0+5(4) \\
=20 \mathrm{~m} / \mathrm{s}
\end{array}
\)
At \(t=6 \mathrm{sec}\)
acceleration is due to gravity
\(
\therefore a=g=10 \mathrm{~m} / \mathrm{s}
\)
\(v_x=20 \mathrm{~m} / \mathrm{s}\) (due to car)
\(
v_y=u+a t
\)
\(
\begin{array}{l}
=0+g(2) \text { (downward) } \\
=20 \mathrm{~m} / \mathrm{s} \text { (downward) } \\
\mathrm{v}=\sqrt{v_x^2+v_y^2} \\
=\sqrt{20^2+20^2} \\
=20 \sqrt{2} \mathrm{~m} / \mathrm{s} .
\end{array}
\)

Hint

Radial acceleration or centripetal accceleration is given by:
\(
a_c=\frac{v^2}{r}=\omega^2 r
\)
Frequency of revolution is, \(f=\frac{n}{t}\) where \(n\) is the number of revolutions, and \(t\) is the time taken to make that revolution.
\(
\begin{array}{l}
f=\frac{22}{44}=\frac{1}{2} \mathrm{~s}^{-1} \\
\Rightarrow \omega=2 \pi f=\pi \mathrm{rad} / \mathrm{s} \\
\therefore a_c=\omega^2 r=\pi^2 \times 1=\pi^2 \mathrm{~m} / \mathrm{s}^2
\end{array}
\)
Tangential acceleration is the rate of change of speed w.r.t time:
\(
\Rightarrow a_t=\frac{d v}{d t}=0
\)
\(\because\) speed is constant
Hence net acceleration of stone \(\vec{a}\) given by:
\(
\begin{array}{l}
\vec{a}=\vec{a}_c+\vec{a}_t=\vec{a}_c=\pi^2 \mathrm{~m} / \mathrm{s}^2 \\
\therefore|\vec{a}|=\left|\vec{a}_c\right|=\pi^2 \mathrm{~m} / \mathrm{s}^2
\end{array}
\)
Direction is along the radius towards the centre i.e. radially inwards.