Motion in a Plane

From graph,

${v_{RG}} = 15\sqrt 2 \tan 45^\circ $

$ = 15\sqrt 2 $

$ = {{30} \over {\sqrt 2 }}$

When two projectiles are thrown with the same initial velocity 'u' but at complementary angles (say, $\theta$ and $(90^\circ - \theta)$) with the horizontal, they attain the same range R. The formula for the range R of a projectile is:

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

For complementary angles, $2 \theta$ and $180^\circ - 2 \theta$ (which simplifies to the same value for the sine function), the ranges are equal.

The maximum height $h_1$ for angle $\theta$ is given by:

$ h_1 = \frac{u^2 \sin^2 \theta}{2g} $

And the maximum height $h_2$ for angle $(90^\circ - \theta)$ is given by:

$ h_2 = \frac{u^2 \cos^2 \theta}{2g} $

Now, multiplying these heights:

$ h_1 h_2 = \left( \frac{u^2 \sin^2 \theta}{2g} \right) \cdot \left( \frac{u^2 \cos^2 \theta}{2g} \right)$

This simplifies to:

$ h_1 h_2 = \frac{u^4 \sin^2 \theta \cos^2 \theta}{4g^2}$

Since $\sin^2 \theta \cos^2 \theta = \left( \frac{\sin 2 \theta}{2} \right)^2 = \frac{1}{4} \sin^2 2 \theta$:

$ h_1 h_2 = \frac{u^4 \sin^2 2 \theta}{16g^2}$

Using the range formula $ R = \frac{u^2 \sin 2 \theta}{g} $, we get:

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

Thus, we have:

$ 4h_1 h_2 = \frac{u^4 \sin^2 2 \theta}{4g^2} = \frac{R^2}{4} $

This simplifies to:

$ R = 4\sqrt{h_1 h_2}$

Both the assertion and the reason are correct, and the reason correctly explains the assertion.

The correct answer is:

Option A : Both A and R are true and R is the correct explanation of A.

$t = {{25\sin \theta } \over g}$

and, $R = {{{{(25)}^2}(2\sin \theta \cos \theta )} \over g}$

$ \Rightarrow R = {{25 \times 25 \times 2} \over g} \times {{gt} \over {25}} \times \cos \theta $

$ \Rightarrow R = 50t\cos \theta $

$\therefore$ $tan\theta = {{gt} \over {25}} \times {{50t} \over R}$

$ = {{20{t^2}} \over R}$

$ \Rightarrow \theta = {\cot ^{ - 1}}\left( {{R \over {20{t^2}}}} \right)$

The given situation is shown below

Velocity of particle $B$ on $Y$-axis,

$ v_B=\frac{d}{d t} y \hat{\mathbf{j}}=\frac{d}{d t} c t^2 \hat{\mathbf{j}}=2 c t \hat{\mathbf{j}} $

As, $c=1 \mathrm{~m} / \mathrm{s}^2$ and $t=1 \mathrm{~s}$

$ v_B(\text { at } t=1 \mathrm{~s})=2 \hat{\mathbf{j}} \mathrm{~m} / \mathrm{s} $

Velocity of particle $A$ along $X$-axis at $t=1 \mathrm{~s}$ is

$ v_A=1 \hat{\mathbf{i}} \mathrm{~m} / \mathrm{s} $

Velocity of $A$ relative to $B$ at $t=1 \mathrm{~s}$ is

$ v_{A B}=\mathbf{v}_A-\mathbf{v}_B=1 \hat{\mathbf{i}}-2 \hat{\mathbf{j}} $

So, speed of $A$ relative to $B=\left|\mathbf{v}_{A B}\right|$

$ =\sqrt{1^2+2^2}=\sqrt{5} \mathrm{~m} / \mathrm{s} $

Velocity component of particle along $X$-axis,

$ \begin{aligned} \mathbf{v}_x & =\frac{d \mathbf{x}}{d t}=\frac{d}{d t}\left(4 t+t^2\right) \hat{\mathbf{i}} \\ & =(4+2 t) \hat{\mathbf{i}} \mathrm{m} / \mathrm{s} \end{aligned} $

Velocity component of particle along $Y$-axis,

$ \mathbf{v}_y=\frac{d}{d t}\left(2 t+\frac{t^2}{2}\right) \hat{\mathbf{j}}=(2+t) \hat{\mathbf{j}} \mathrm{m} / \mathrm{s} $

So, velocity of particle is

$ \begin{aligned} \mathbf{v} & =\mathbf{v}_x+\mathbf{v}_y \\ & =(4+2 t) \hat{\mathbf{i}}+(2+t) \hat{\mathbf{j}} \mathrm{m} / \mathrm{s} \end{aligned} $

$ \text { Given situation is shown below. } $

As it is clear from above diagram, velocity perpendicular to incline

$ \begin{aligned} & =x \cos 30^{\circ} \text { or } u_{\perp}=u \cos 30^{\circ}(\text { upward }) \\ & =10 \times \frac{\sqrt{3}}{2}=5 \sqrt{3} \mathrm{~ms}^{-1} \end{aligned} $

And acceleration perpendicular to inclined plane is

$ \begin{aligned} a_{\perp} & =-g \cos 30^{\circ} \text { or }-g \sin 60^{\circ} \\ & =-g \times \frac{\sqrt{3}}{2}(\text { downwards }) \\ & =-5 \sqrt{3} \mathrm{~m} / \mathrm{s}^2 \end{aligned} $

Let projected particle again lands on incline in time $t$, then

So, times of flight $t$ is given by

$ \begin{aligned} & s=u_{\perp} t+\frac{1}{2} a_{\perp} t^2 \\ \Rightarrow & 0=5 \sqrt{3} t-\frac{5 \sqrt{3}}{2} \times t^2 \\ \Rightarrow & t=0 \text { or } t=2 \mathrm{~s} \end{aligned} $

In two seconds particle lands on the incline.

Initial velocity,

$ x_{\|}=x \cos 60^{\circ}=\frac{x}{2} \mathrm{~m} / \mathrm{s}=5 \mathrm{~m} / \mathrm{s} $

acceleration down inline,

$ a_{\|}=g \cos 60^{\circ}=\frac{g}{2} \mathrm{~m} / \mathrm{s}^2=5 \mathrm{~m} / \mathrm{s}^2 $

Now, using $s=u t+\frac{1}{2} a t^2$

We get,

$ \begin{aligned} s & =5 \times 2+\frac{1}{2} \times 5 \times 2^2 \\ & =20 \mathrm{~m} \end{aligned} $

So, particle covers 20 m down the plane.

When an object has two velocities $\mathbf{v}_1$ and $\mathbf{v}_2$, then magnitude of resultant velocity is

$ \mathbf{v}=\sqrt{v_1^2+v_2^2+2 v_1 v_2 \cos \theta} $

Statement I is incorrect.

Clearly | displacement| $\leq$ distance

Statement II is correct.

$ \text { Instantaneous acceleration }=\lim\limits_{\Delta t \rightarrow 0} \frac{\Delta \mathbf{v}}{\Delta t}=\lim\limits_{\Delta t \rightarrow 0} \mathbf{a} $

Statement III is correct.

The given situation is shown below

For a projectile, Range, $\quad R=\frac{u^2 \sin 2 \alpha}{g}$

Maximum height, $\quad h=\frac{u^2 \sin ^2 \alpha}{2 g}$

Time of flight,           $t=\frac{2 u \sin \alpha}{g}$

$ \begin{aligned} & \text { } \begin{aligned} Now, \frac{h}{R} & =\frac{u^2 \sin ^2 \alpha / 2 g}{u^2 \sin 2 \alpha / g}=\frac{\sin ^2 \alpha}{2 \sin 2 \alpha} \\ & =\frac{\sin ^2 \alpha}{2 \cdot 2 \sin \alpha \cos \alpha}(\because \sin 2 \theta=2 \sin \theta \cdot \cos \theta)=\frac{1}{4} \tan \alpha \\ \Rightarrow \quad \tan \alpha & =\frac{4 h}{R} \\ \text { Also, } \quad \frac{h}{t^2} & =\frac{u^2 \sin ^2 \alpha / 2 g}{4 u^2 \sin ^2 \alpha / g^2} \Rightarrow \frac{h}{t^2}=\frac{g}{8} \text { or } h=\frac{g t^2}{8} \end{aligned} \end{aligned} $

The given situation is shown below.

The two cars $A$ and $B$ are moving with same speed as $25 \mathrm{~km} / \mathrm{h}$ in the direction shown in the above figure.

From the diagram,

Relative velocity of car $B$ with respect to car A,

$ v_{B A}=v_B-v_A=\sqrt{25^2+25^2}=25 \sqrt{2} \mathrm{~km} / \mathrm{h} $

i.e. $v_{B A}$ is $25 \sqrt{2} \mathrm{~km} / \mathrm{h}$ at an angle of $45^{\circ}$ from east towards north.

∴ Minimum distance between the two cars i.e distance of closest approach is equal

$ \begin{aligned} A C & =B A \sin 45^{\circ} . \\ & =50 \sin 45^{\circ}=25 \sqrt{2} \mathrm{~km} \\ \therefore \text { Time taken } & =\frac{A C}{\left|v_{\mathrm{BA}}\right|}=\frac{25 \sqrt{2}}{25 \sqrt{2}}=1 \mathrm{~h} \\ & =60 \mathrm{~min} \end{aligned} $

Given, $\mathbf{v}=(6+2 t) \hat{\mathbf{i}}+(4+2 \sqrt{3} t) \hat{\mathbf{j}}$

Comparing with, $v=v_x \hat{\mathbf{i}}+v_y \hat{\mathbf{j}}$

We get, $v_x=6+2 t$ and $v_y=4+2 \sqrt{3} t$

$ \therefore \quad a_x=\frac{d}{d t} v_x=\frac{d}{d t}(6+2 t)=2 \mathrm{~ms}^{-1} $

$ \begin{aligned}and\,\,\, a_y & =\frac{d}{d t} v_y=\frac{d}{d t}(4+2 \sqrt{3} t) \\ & =2 \sqrt{3} \mathrm{~ms}^{-2} \\\therefore \,\,\,\,\, a & =a_x \hat{\mathbf{i}}+a_y \hat{\mathbf{j}}=2 \hat{\mathbf{i}}+2 \sqrt{3} \hat{\mathbf{j}} \end{aligned} $

The given situation is shown below.

At time $t=0$,

$ \begin{aligned} u_x & =u \cos \theta=20 \cos 30^{\circ} \\ & =20 \frac{\sqrt{3}}{2}=10 \sqrt{3} \mathrm{~m} / \mathrm{s} \\ u_y & =u \sin \theta=20 \sin 30^{\circ} \\ & =20 \times \frac{1}{2}=10 \mathrm{~m} / \mathrm{s} \end{aligned} $

Let after time $t=0.1 \mathrm{~s}$, fired bullet reaches at point $A$. If $v$ be the velocity and $\alpha$ be the angle made by displacement vector from horizontal at point $A$, then

$ \begin{aligned} & v \cos \alpha=u_x=10 \sqrt{3} \\ \Rightarrow \quad & v \cos \alpha=10 \sqrt{3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{aligned} $

and $v \sin \alpha=u \sin \theta-g t$

$ \begin{aligned} & =u_y-g t \\ & =10-10 \times 0.1 \\ \Rightarrow \quad v \sin \alpha & =9 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Dividing Eq. (ii) by Eq. (i), we get

$ \begin{aligned} & \frac{v \sin \alpha}{v \cos \alpha} =\frac{9}{10 \sqrt{3}} \\ \Rightarrow \quad & \tan \alpha =\frac{9}{10 \sqrt{3}} \end{aligned} $

$ \text { The given situation is shown below } $

If $\theta$ be the angle of umbrella from vertical to protect himself from rain, then

$ \tan \theta=\frac{v_{\text {man }}}{v_{\text {rain }}}=\frac{6}{6 \sqrt{3}}=\frac{1}{\sqrt{3}}=\tan 30^{\circ} $

$\Rightarrow \quad \theta=30^{\circ}$

Initial velocity,

$ \begin{aligned} & \mathbf{u}=(3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s} \\ & \tan \theta=\frac{4}{3} \\ & \therefore \quad \theta=53^{\circ} \end{aligned} $

Equation of trajectory $\Rightarrow \frac{1}{9}\left[\beta x+\gamma x^2\right]$

We know that, equation of trajectory of a projectile is

$ y=x \tan \theta-\frac{g x^2}{2 u^2 \cos ^2 \theta} $

On Comparing with $\frac{1}{9} \beta x+\frac{\gamma}{9} x^2$

We find that,

$ \begin{aligned} \frac{\beta}{9} & =\tan \theta \text { and } \frac{\gamma}{9}=\frac{-g}{2 u^2 \cos ^2 \theta} \\ \Rightarrow \frac{\beta}{9} & =\tan 53^{\circ} \text { and } \frac{\gamma}{9}=\frac{-10}{2 u^2 \cos ^2 53^{\circ}} \\ \Rightarrow \beta & =9 \times \frac{4}{3}=12 \text { and } \frac{\gamma}{9}=\frac{-5}{u^2 \times\left(\frac{3}{5}\right)^2} \end{aligned} $

$ \left(\because \cos 53^{\circ}=\frac{3}{5}\right) $

$\therefore r=\frac{-9 \times 5 \times 25}{u^2 \times 9} \Rightarrow r=\frac{-125}{u^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

$ \begin{array}{ll} \text { Now, } & \because \mathbf{u}=(3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s} \\ \therefore & |\mathbf{u}|=\sqrt{3^2+4^2}=5 \mathrm{~m} / \mathrm{s} \end{array} $

$ \begin{array}{ll} \therefore r=\frac{-125}{5^2} \quad \text { [from Eq. (i)] } \\ r =\frac{-125}{25}=-5 \end{array} $

The given situation is shown below.

Let the speed at point $P=v$

By the conservation of energy,

$ \therefore \quad-\Delta U=\Delta K $

where, $-\Delta U=$ decrease in potential energy and $\Delta K=$ increase in kinetic energy of object.

$ \begin{array}{ll} \Rightarrow & -\left(U_f-U_i\right)=K_f-K_i \\ \Rightarrow & -\left(m g \frac{H}{2}-m g H\right)=\frac{1}{2} m v^2-\frac{1}{2} m u^2 \\ \Rightarrow & m g \frac{H}{2}=\frac{1}{2} m v^2-0 \\ \Rightarrow & v=\sqrt{g H} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{array} $

Now, when the object reaches point $P$, let its speed be $v$ in horizontal direction and $R$ be the distance from the end of the hill to the ground and also the time

aken by the object to travel from $P$ to $Q$

$ \begin{array}{r} \therefore y=u_y t+\frac{1}{2} a_y t^2 \Rightarrow \frac{H}{2}=0+\frac{1}{2} g t^2 \\ \left(\because u_y=0 \text { at } P\right) \end{array} $

$ \Rightarrow t=\sqrt{\frac{2 H}{2 g}} \Rightarrow t=\sqrt{\frac{H}{g}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

and $x=u_x t+\frac{1}{2} a_x t^2$

$ \Rightarrow R=v t+0 \,\,\,\,\,\,\left(\because a_x=0\right)$

$ \begin{aligned} &\Rightarrow R=v t \quad \Rightarrow R=\sqrt{g H} \sqrt{\frac{H}{g}}\\ &\text { [From Eqs. (i) and (ii)] }\\ &\Rightarrow \quad R=\sqrt{\frac{g H^2}{g}}=H \quad \Rightarrow R=H \end{aligned} $

The given situation is shown below

Since, time of flight, $T=\frac{2 u \sin \theta}{g}$

$ =\frac{2 \times 40 \times \sin 30^{\circ}}{10}=\frac{2 \times 40 \times 1}{10 \times 2}=4 \mathrm{~s} $

∴ In 2 s , it will be at the topmost point, where it will hit the hillside.

After 2 s , the maximum height ( $H_{\max }$ )

$ \begin{aligned} A B & =H_{\max }=\frac{u^2 \sin ^2 \theta}{2 \mathrm{~g}} \\ & =\frac{40^2 \sin ^2 30^{\circ}}{2 \times 10}=\frac{1600 \times 1}{2 \times 10 \times 4}=20 \mathrm{~m} \end{aligned} $

Also, at this time the horizontal distance travelled will be half of the range $=O A$

$ \text { } \begin{aligned} i.e\frac{R}{2} O A & =\frac{R}{2}=\frac{u^2 \sin 2 \theta}{2 g}=\frac{40^2 \sin \left(2 \times 30^{\circ}\right)}{2 \times 10} \\ & =\frac{1600 \sin 60^{\circ}}{20} \end{aligned} $

$ \therefore \quad \frac{R}{2}=80 \frac{\sqrt{3}}{2}=40 \sqrt{3} $

∴ The displacement from the projection point,

$ O B=D=\sqrt{O A^2+A B^2} $

(By Pythagorous theorem)

$ \begin{aligned} & \Rightarrow D=\sqrt{(40 \sqrt{3})^2+(20)^2}=\sqrt{4800+400} \\ & =\sqrt{5200} \\ & \Rightarrow D=\sqrt{4 \times 13 \times 100}=2 \times 10 \sqrt{13} \\ & =20 \sqrt{13} \mathrm{~m} \end{aligned} $

To solve this problem, we will use the kinematic equations for projectile motion. Let's assume the projectile is launched at an angle $\theta$ with an initial velocity $v_0$. The range of the projectile, which is the horizontal distance it covers, is given by:

$R = \frac{{v_0^2 \sin 2\theta}}{{g}}$

where:

• $R$ is the range (90 meters in this case)
• $v_0$ is the initial velocity
• $g$ is the acceleration due to gravity (10 $\mathrm{~ms}^{-2}$)
• $\theta$ is the angle of projection

To hit the target at the minimum velocity, the projectile should be launched at the optimum angle, which is 45 degrees ($\sin 2\theta = \sin 90^\circ = 1$). Thus, the equation simplifies to:

$R = \frac{{v_0^2}}{{g}}$

Now, rearrange the equation to solve for $v_0$:

$v_0^2 = R \cdot g$

Substitute the values of $R$ and $g$:

$v_0^2 = 90 \, \mathrm{m} \cdot 10 \, \mathrm{ms}^{-2}$

$v_0^2 = 900 \, \mathrm{m^2s}^{-2}$

Take the square root of both sides to find $v_0$:

$v_0 = \sqrt{900} \, \mathrm{ms}^{-1}$

$v_0 = 30 \, \mathrm{ms}^{-1}$

Therefore, the minimum velocity of the projectile to hit the target is:

Option D: $30 \mathrm{~ms}^{-1}$

The given situation is shown below

Given, $m=1 \mathrm{~kg}$

$\begin{aligned} & \left|F_1\right|=4 \mathrm{~N} \\ & \left|F_2\right|=4 \mathrm{~N} \end{aligned}$

From the figure, angle between force, $F_1$ and $F_2, \theta=60^{\circ}$

$\because$ Magnitude of resultant force,

$\begin{aligned} F & =\sqrt{F_1^2+F_2^2+2 F_1 F_2 \cos \theta} \\ & =\sqrt{4^2+4^2+2 \times 4 \times 4 \times \cos 60^{\circ}} \\ & =\sqrt{16+16+16}=4 \sqrt{3} \mathrm{~N} \end{aligned}$

Magnitude of acceleration of the object

$\begin{aligned} \mathbf{a} & =\frac{F}{m}=\frac{4 \sqrt{3}}{1}=4 \sqrt{3} \\ & =6.9 \mathrm{~m} / \mathrm{s}^2 \end{aligned}$

The given situation can be described by the figure given below

$\begin{aligned} & \begin{aligned} \text { Thus, } & \sin \theta=\frac{40}{\text { Velocity of rain w.r.t car }} \\ & \Rightarrow \text { velocity of rain }=\frac{40}{\sin \theta} \\ &=\frac{40}{\frac{1}{2}} \quad\left[\because \theta=30^{\circ}\right]\\ &=40 \times 2=80 \mathrm{~kmh}^{-1} \end{aligned} \end{aligned}$

Speed of projectile, $u=50 \mathrm{~ms}^{-1}$

Angle of projection,

$\begin{aligned} & \theta=60^{\circ} \\ & g=10 \mathrm{~ms}^{-2} \end{aligned}$

Maximum height attained by projectile,

$\begin{aligned} H & =\frac{u^2 \sin ^2 \theta}{2 g} \\ & =\frac{50^2 \times \sin ^2 60^{\circ}}{2 \times 10}=\frac{2500 \times\left(\frac{\sqrt{3}}{2}\right)^2}{20} \\ & =125 \times \frac{3}{4}=\frac{375}{4} \\ & =93.75 \mathrm{~m} \end{aligned}$

The correct answer is Option C, a straight line vertically down the plane.

Here's why:

• Frame of Reference: The key to understanding projectile motion is considering the frame of reference. The observer in the plane is in a moving frame of reference. From their perspective, they are at rest, and the bomb has the same horizontal velocity as the plane.


• Gravity's Influence: The only force acting on the bomb after it's released is gravity. Gravity acts vertically downwards.


• Resultant Motion: Because the bomb has the same horizontal velocity as the plane and is only affected by gravity vertically, it appears to the observer in the plane to fall straight down.

Let's eliminate the other options:

• Option A (Hyperbola): A hyperbolic trajectory occurs when an object is influenced by two forces acting in different directions, resulting in a path that gets progressively straighter. This doesn't apply to the bomb scenario.


• Option B and D (Parabola): A parabolic trajectory is observed from a stationary frame of reference on the ground. In this case, the horizontal motion of the bomb combined with the vertical acceleration due to gravity creates a parabolic path.

Given, height of cliff, $H=490 \mathrm{~m}$

Initial speed along $X$-axis, $u_x=15 \mathrm{~ms}^{-1}$

and initial speed along $Y$-axis, $u_y=0 \mathrm{~ms}^{-1}$

Acceleration due to gravity $g=10 \mathrm{~ms}^{-2}$

As we know that, for motion along $Y$-axis,

$\begin{aligned} v_y^2-u_y^2 & =2 g H \\ \Rightarrow \quad v_y & =\sqrt{2 g H} \\ & =\sqrt{2 \times 10 \times 490}=70 \sqrt{2} \mathrm{~ms}^{-1} \\ & =99 \mathrm{~ms}^{-1} \end{aligned}$

Given, separation between two papers $A$ and $B, s=150 \mathrm{~m}$

Height between hole $A$ and $B, H=15 \mathrm{~cm}$

$=15 \times 10^{-2} \mathrm{~m}$

Acceleration due to gravity, $g=10 \mathrm{~ms}^{-2}$

Let, time taken to go from $A$ to $B$ be $t$.

Velocity of bullet at point $A$ be $v$.

Since, $H=(1/2)gt^2$

$\begin{array}{ll} \Rightarrow & t=\sqrt{\frac{2 H}{g}} \quad \text { [for accelerated motion] } \\ \Rightarrow \quad & t=\sqrt{\frac{2 \times 15}{100 \times 10}}=10^{-1} \sqrt{3} \mathrm{~s} \end{array}$

and $v=\frac{s}{t}$ [for non-accelerated motion] $ \therefore \quad v=\frac{150}{10^{-1} \sqrt{3}}=500 \sqrt{3} \mathrm{~ms}^{-1} $

Given, friction force due to air = $f_a$

According to given diagram

Let W be the weight of ball.

As, we know that, weight W is always perpendicular towards earth and friction force $f_a$ is always in opposite direction of net force.

$\therefore$ Free body diagram of ball will be

Given, mass of particle $=m$

Velocity $=u$

Angle of projection $=\theta$

Let angular momentum, $L=m v r$

Maximum height $=H$

As we know that,

$\begin{aligned} & H=\frac{u^2 \sin ^2 \theta}{2 g}=r \\ & \therefore \quad L=m \cdot u \cos \theta \cdot r \\ & \Rightarrow \quad L=m u \cos \theta\left(\frac{u^2 \sin ^2 \theta}{2 g}\right) \\ & \Rightarrow \quad L=\frac{m u^3 \sin ^2 \theta \cos \theta}{2 g} \\ \end{aligned}$

The time of flight of a projectile is the time it takes for the projectile to return to the same vertical position from which it was launched. The time of flight can be calculated using the following equation:

$T = \frac{2u\sin\theta}{g}$

where:

T = time of flight

u = initial velocity

θ = angle of projection

g = acceleration due to gravity

In this case, the initial velocity is 50 ms$^{-1}$, the angle of projection is 30$\Upsilon$, and the acceleration due to gravity is 10 ms$^{-2}$. Plugging these values into the equation, we get:

$T = \frac{2 \times 50 \times \sin30}{10}$

$T = \frac{2 \times 50 \times 0.5}{10}$

$T = \frac{50}{10}$

$T = 5 s$

Therefore, the ball remains in the air for 5 s.

So the answer is Option A.

Given, $m=1 \mathrm{~kg}$

Force, $F(t)=[2 \sin \alpha t \hat{\mathbf{i}}+3 \cos \alpha t \hat{\mathbf{j}}] \mathrm{N}$

Here, $\alpha=1 \mathrm{~s}^{-1}$

$ \Rightarrow \quad F(t)=(2 \sin t \hat{\mathbf{i}}+3 \cos t \hat{\mathbf{j}}) \mathrm{N} $

So, component of force,

$ F_x=2 \sin t \text { and } F_y=3 \cos t $

As, $\quad F_x=\frac{m d v_x}{d t}=2 \sin t$

$ \Rightarrow \quad d v_x=2 \sin t d t \,\,\,\,\,\,\,(\because m=1 \mathrm{~kg}) $

Integrating both sides, we get

$ \begin{aligned} & v_x=\int_0^{\pi / 2} 2 \sin t d t=-2[\cos t]_0^{\pi / 2} \\ & =-2\left(\cos \frac{\pi}{2}-\cos 0\right)=-2(0-1)=2 \mathrm{~m} / \mathrm{s} \end{aligned} $

Similarly, $v_y=\int_0^{\pi / 2} 3 \cos t d t=3[\sin t]_0^{\pi / 2}$

$ =3\left(\sin \frac{\pi}{2}-\sin 0\right)=3(1-0)=3 \mathrm{~m} / \mathrm{s} $

∴ Magnitude of velocity vector,

$ |v|=\sqrt{v_x^2+v_y^2}=\sqrt{(2)^2+(3)^2}=\sqrt{13} \mathrm{~m} / \mathrm{s} $

As, velocity vector,

$ \mathbf{v}=v_x \hat{\mathbf{i}}+v_y \hat{\mathbf{j}}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}} $

So, $\quad v_x=\frac{d r_x}{d t}=2 \mathrm{~m} / \mathrm{s}$ and

$ \begin{aligned} v_y & =\frac{d r_y}{d t}=3 \mathrm{~m} / \mathrm{s} \\ \Rightarrow \quad d r_x & =2 d t \end{aligned} $

Integrating both sides, we get

$ r_x=\int_0^{\pi / 2} 2 d t=2[t]_0^{\pi / 2}=\pi \mathrm{m} $

Similarly, $r_y=\int_0^{\pi / 2} 3 d t=3[t]_0^{\pi / 2}=\frac{3 \pi}{2} \mathrm{~m}$

∴ Magnitude of position vector,

$ \begin{aligned} |r| & =\sqrt{r_x^2+r_y^2}=\sqrt{\pi^2+\left(\frac{3 \pi}{2}\right)^2} \\ & =\sqrt{13} \frac{\pi}{2} \mathrm{~m} \end{aligned} $

Let $d$ be the distance of target.

The horizontal range of projectile is

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

When $\theta=15^{\circ}$,

$ R_1=\frac{u^2 \sin \left(2 \times 15^{\circ}\right)}{g}=\frac{u^2}{2 g} $

When $\theta=45^{\circ}$,

$ R_2=\frac{u^2 \sin \left(2 \times 45^{\circ}\right)}{g}=\frac{u^2}{g} $

According to question,

$ \begin{equation*} R_1=\frac{u^2}{2 g}=d-10\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{equation*} $

$ \begin{equation*} \text { and } R_2=\frac{u^2}{g}=d+10 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{equation*} $

Dividing Eq. (i) by Eq. (ii), we get

$ \begin{array}{rlrl} & & \frac{1}{2} & =\frac{d-10}{d+10} \\ \Rightarrow & d & =30 \mathrm{~m} \tag{iii} \end{array} $

From Eq. (ii), $\frac{u^2}{g}=30+10=40\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

So, to hit the target at 30 m ,

$ \begin{aligned} 30 & =\frac{u^2 \sin 2 \theta}{g} \Rightarrow 30=40 \sin 2 \theta\,\,\,\,\,\,\, \text { [from Eq. (iii)] }\\ \Rightarrow \quad \theta & =\frac{1}{2} \sin ^{-1}\left(\frac{3}{4}\right) \end{aligned} $

The given situation is shown in the following figure

Speed of car $A$,

$ v_A=120 \mathrm{~km} / \mathrm{h} \Rightarrow \mathbf{v}_A=-120 \hat{\mathbf{i}} $

Speed of car $B$,

$ \begin{aligned} & v_B=50 \mathrm{~km} / \mathrm{h} \\ & \mathbf{v}_B=-50 \hat{\mathbf{j}} \end{aligned} $

∴ Relative speed of the car $B$ w.r.t. car $A$,

$ \begin{aligned} \mathbf{v}_{B A} & =\mathbf{v}_B-\mathbf{v}_A=-50 \hat{\mathbf{j}}-(-120 \hat{\mathbf{i}}) \\ & =-50 \hat{\mathbf{j}}+120 \hat{\mathbf{i}}=120 \hat{\mathbf{i}}-50 \hat{\mathbf{j}} \\ \therefore \quad\left|\mathbf{v}_{B A}\right| & =\sqrt{(120)^2+(-50)^2} \\ & =\sqrt{16900}=130 \mathrm{~km} / \mathrm{h} \end{aligned} $

Initial velocity of projectile, $u=10 \mathrm{~m} / \mathrm{s}$

Angle of projection from inclined plane, $\theta=60^{\circ}-30^{\circ}=30^{\circ}$

Here, $\alpha=30^{\circ}$

Range over an inclined plane is given as

$ \begin{aligned} R & =\frac{2 u^2}{g} \cdot \frac{\sin \theta \cos (\theta+\alpha)}{\cos ^2 \alpha} \\ & =\frac{2 \times 10^2}{10} \times \frac{\sin 30^{\circ} \cos \left(30^{\circ}+30^{\circ}\right)}{\cos ^2 30^{\circ}} \\ & =20 \times \frac{\frac{1}{2} \times \frac{1}{2}}{\left(\frac{\sqrt{3}}{2}\right)^2}=20 \times \frac{\frac{1}{3}}{\frac{3}{4}}=\frac{20}{3} \mathrm{~m} \end{aligned} $

The given situation is shown in the following figure

Angle of projection, $\theta=45^{\circ}$

The elevation angle of the projectile at its highest point as seen from the point of projection is $\alpha$.

∴ In $\triangle P O M$,

$ \tan \alpha=\frac{H}{R / 2}=\frac{\frac{u^2 \sin ^2 \theta}{2 g}}{\frac{u^2 \sin 2 \theta}{2 g}}=\frac{\sin ^2 \theta}{\sin 2 \theta} $

$ \begin{aligned} & =\frac{\sin ^2 45^{\circ}}{\sin 90^{\circ}} \\ & =\frac{\left(\frac{1}{\sqrt{2}}\right)^2}{1} \\ \Rightarrow \quad \tan \alpha & =\frac{1}{2} \\ \Rightarrow \quad \alpha & =\tan ^{-1}\left(\frac{1}{2}\right) \end{aligned} $