Motion in a Plane

$y=\sqrt{3} x-0.2 x^2$

Comparing with equation of trajectory.

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

We get, $\tan \theta=\sqrt{3}=\tan 60^{\circ}$

$ \begin{aligned} & \therefore \quad \theta=60^{\circ} \\ & \therefore \quad \frac{g}{2 u^2 \cos ^2 \theta}=0.2 \\ & \Rightarrow \frac{10}{2 u^2 \cos ^2 60^{\circ}}=0.2 \\ & \Rightarrow \frac{5}{u^2 \times \frac{1}{4}}=0.2 \Rightarrow \frac{20}{u^2}=0.2 \\ & \Rightarrow \quad u^2=\frac{20}{0.2}=100 \\ & \therefore \quad u=10 \mathrm{~m} / \mathrm{s} \\ & \therefore \text { Time of flight, } T=\frac{2 u \sin \theta}{g} \\ & \quad=\frac{2 \times 10 \times \sin 60^{\circ}}{10}=\sqrt{3} \mathrm{~s} \end{aligned} $

Step 1: Find the total time the object is in the air

The question says the object crosses a certain point after 2.3 s, then takes 5.7 s from that point to reach the ground. So, the total time in the air is:

$T = 2.3~\mathrm{s} + 5.7~\mathrm{s} = 8~\mathrm{s}$

Step 2: Find the time taken to reach the maximum height

For a projectile (an object thrown at an angle), the time to go up to the highest point is half of the total time in the air:

$t_{\text{max}} = \frac{T}{2} = \frac{8~\mathrm{s}}{2} = 4~\mathrm{s}$

Step 3: Find the starting upward speed using the motion formula

At the highest point, the speed going up becomes zero. The formula is:

$v_f = v_i - g t_{\text{max}}$

Here, $v_f = 0$ (since upward speed is zero at the top), $g = 10~\mathrm{m/s}^2$, and $t_{\text{max}} = 4~\mathrm{s}$:

$0 = v_i - (10 \times 4)$ $v_i = 40~\mathrm{m/s}$

Step 4: Calculate the maximum height reached

We use another formula to find the highest point:

$ H_{\text{max}} = v_i t_{\text{max}} - \frac{1}{2} g t_{\text{max}}^2 $

Plug in the values:

$ H_{\text{max}} = (40~\mathrm{m/s})(4~\mathrm{s}) - \frac{1}{2} (10~\mathrm{m/s}^2)(4~\mathrm{s})^2 $

$(40 \times 4) = 160~\mathrm{m}$

$(4~\mathrm{s})^2 = 16~\mathrm{s}^2$

$\frac{1}{2} \times 10 \times 16 = 80~\mathrm{m}$

$H_{\text{max}} = 160~\mathrm{m} - 80~\mathrm{m} = 80~\mathrm{m}$

Final Answer:The maximum height reached by the body is 80 m.

Time of flight of projectile

$ (T)=2+8=10 \mathrm{~s} $

Horizontal distance between two points

$ \begin{aligned} = & u \cos 45^{\circ} \times 8-u \cos 45^{\circ} \times 2 \\ = & 6 u \cos 45^{\circ} \\ T & =\frac{2 u \sin \theta}{g}=\frac{2 u \sin 45^{\circ}}{10} \\ 10 & =\frac{2 u \sin 45^{\circ}}{10} \\ u & =50 \sqrt{2} \mathrm{~m} / \mathrm{s} \end{aligned} $

$ \begin{aligned} &\text { Horizontal distance between two points }\\ &=6 \times 50 \sqrt{2} \times \frac{1}{\sqrt{2}}=300 \mathrm{~m} \end{aligned} $

Step 1: Write the formula for maximum height

The formula for the highest point a projectile can reach is:

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

where $H$ is maximum height, $u$ is the starting speed, $\theta$ is the angle of projection, and $g$ is gravity.

Step 2: Put in the given values

We know $H = 9.8$ m, $u = 19.6~\mathrm{ms}^{-1}$, and $g = 9.8~\mathrm{ms}^{-2}$. We put these in the formula:

$ 9.8 = (19.6)^2 \frac{\sin^2 \theta}{2 \times 9.8} $

Step 3: Simplify to find $\sin^2\theta$

Multiply both sides by $2 \times 9.8$:

$ 9.8 = 19.6 \times 19.6 \times \frac{\sin^2 \theta}{2 \times 9.8} $

$ \Rightarrow 9.8 = 39.2 \sin^2 \theta $

Divide both sides by $19.6$,

$ \sin^2\theta = \frac{1}{2} $

Step 4: Find the angle $\theta$

$\sin\theta = \frac{1}{\sqrt{2}} = \sin 45^\circ$, so

$\theta = 45^\circ$.

Step 5: Write the formula for range $R$

The formula for how far the projectile lands (range) is:

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

Step 6: Put in the numbers

Since $u = 19.6~\mathrm{ms}^{-1}$, $\theta = 45^\circ$, and $g = 9.8~\mathrm{ms}^{-2}$,

$ R = \frac{(19.6)^2 \times \sin(2 \times 45^\circ)}{9.8} $

$\sin(90^\circ) = 1$, so

$ R = \frac{(19.6)^2 \times 1}{9.8} $

$ R = \frac{384.16}{9.8} = 39.2~\mathrm{m} $

The given situation is shown below in figure.

$ x_1=\frac{R_1}{2}=\frac{u^2 \sin 2 \theta}{2 g} $

and $x_2=\frac{R_2}{2}=\frac{u^2 \sin 2\left(90^{\circ}-\theta\right)}{2 g}$

$ \therefore \quad=\frac{u^2 \sin \left(180^{\circ}-2 \theta\right)}{2 g}=\frac{u^2 \sin 2 \theta}{2 g} $

$\therefore d=x_1+x_2$

$ \begin{aligned} & =\frac{u^2 \sin 2 \theta}{2 g}+\frac{u^2 \sin 2 \theta}{2 g}=\frac{u^2 \sin 2 \theta}{g} \\ & =\frac{u^2 \sin \left(180^{\circ}-2 \theta\right)}{g}=\frac{u^2 \sin 2\left(90^{\circ}-\theta\right)}{g} \end{aligned} $

$ \text { } \begin{aligned} u & =288 \mathrm{~km} / \mathrm{h} \\ & =288 \times \frac{5}{18}=80 \mathrm{~m} / \mathrm{s} \\ h & =\frac{1}{2} g t^2 \\ x & =u t \end{aligned} $

$ \begin{array}{ll} \therefore & \tan 45^{\circ}=\frac{h}{x} \\ \Rightarrow & h=x \Rightarrow \frac{1}{2} g t^2=u t \\ \Rightarrow & t=\frac{2 u}{g}=\frac{2 \times 80}{10} \Rightarrow t=16 \mathrm{~s} \end{array} $

$ \begin{aligned} &\therefore \text { Height of bomb, } h=\frac{1}{2} g t^2\\ &=\frac{1}{2} \times 10 \times 16^2=1280 \mathrm{~m} \end{aligned} $

The maximum horizontal range $ (R) $ of a projectile launched at an optimal angle $ \left(45^{\circ}\right) $ is given by,

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

Given, $ R=32 \mathrm{~m} $

Then $ 32=\frac{u^{2} \cdot \sin 2 \cdot 45^{\circ}}{10} $

$ \begin{array}{l} u^{2}=320 \\\\ u=8 \sqrt{5} \mathrm{~m} / \mathrm{s} \end{array} $

When the ball is thrown horizontally from the top of a tower of height 25 m , the vertical motion is given by,

$ y=\frac{1}{2} g t^{2} $

Given, $ y=25 \mathrm{~m} $

Then, $ 25=\frac{1}{2} \times 10 \times t^{2} $

$ t=\sqrt{5} \mathrm{~s} $

The horizontal distance $ x $ covered by the ball is given by,

$ x=u t $

From Eqs. (i) and (ii), we get

$ \begin{array}{l} x=8 \sqrt{5} \times \sqrt{5} \\\\ x=8 \times 5 \\\\ x=40 \mathrm{~m} \end{array} $

Initial speed, $ u=40 \mathrm{~m} / \mathrm{s} $

Final speed, $ v=1.25 \times u $

$ v=50 \mathrm{~m} / \mathrm{s} $

$ \because $ Horizontal component is constant

$ \begin{array}{l} \Rightarrow \quad 40 \times \frac{\sqrt{3}}{2}=50 \cos \phi \\\\ \Rightarrow \quad \cos \phi=\frac{4 \sqrt{3}}{10} \end{array} $

Vertical component

$ \begin{aligned} v_{y} & =v \sin \phi=v \sqrt{1-\cos ^{2} \phi} \\\\ v_{y} & =50 \sqrt{1-\left(\frac{4 \sqrt{3}}{10}\right)^{2}} \\\\ & =5 \sqrt{52} \end{aligned} $

Now using 3rd equation of motion

$ \begin{aligned} \Rightarrow \quad & (5 \sqrt{52})^{2}-\left(40 \sin 30^{\circ}\right)^{2}=2 g h \\\\ \Rightarrow \quad & h=\frac{1300-400}{2 \times 10}=45 \mathrm{~m} \\\\ & h=45 \mathrm{~m} \end{aligned} $

Given,

$\theta=\tan ^{-1}(\sqrt{7})$

Maximum height

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

[Put $ \theta=\tan ^{-1}(\sqrt{7}) $ ]

$ \begin{array}{l} H_{\max }=\frac{7}{8} \frac{u^{2}}{2 g} \\\\ \frac{H_{\max }}{2}=\frac{7}{16} \frac{u^{2}}{2 g} \end{array} $

Using 3rd equation of motion for velocity at $ \frac{H_{\text {max }}}{2} $

$ v^{2}=u^{2}-2 g h $

where $ v=n u $ (given)

$ \begin{aligned} h & =\frac{H_{\max }}{2} \\\\ n^{2} u^{2} & =u^{2}-2 g \frac{H_{\max }}{2} \\\\ \left(n^{2}-1\right) u^{2} & =-2 g \times \frac{7}{16} \frac{u^{2}}{2 g} \\\\ n^{2}-1 & =\frac{-7}{16} \Rightarrow n^{2}=\frac{-7}{16}+1 \\\\ n^{2} & =\frac{9}{16} \\\\ n & =\frac{3}{4} \end{aligned} $

Given, initial height, $y_i=375 \mathrm{~m}$

Final height, $y_f=0 \mathrm{~m}$

Initial velocity, $v_i=100 \mathrm{~m} / \mathrm{s}$

Initial horizontal range, $x_i=0$

Angle of projection, $\theta=30^{\circ}$

Now, $y_f=y_i+v_i t-\frac{1}{2} g t^2$

$ \begin{aligned} & y_f=y_i+v_i g \sin \theta t-\frac{1}{2} g t^2 \\\\ & 0=375+\left(100 \times \sin 30^{\circ}\right) t-\frac{1}{2} \times 9.8 t^2 \\\\ & 0=375+50 t-4.9 t^2 \\\\ & \Rightarrow 4.9 t^2-50 t-375=0 \end{aligned} $

On solving, we get

$ t_1=15.22 \mathrm{~s} \text { and } t_2=-5.02 \mathrm{~s} $

Now time cannot be negative hence,

$ t=15.22 \mathrm{~s} $

Thus horizontal rang, $R=v_t \cos \theta t$

$ \begin{aligned} & =100 \times \cos 30^{\circ} \times 15.22 \\\\ & =100 \times \frac{\sqrt{3}}{2} \times 15.22 \\\\ & =761 \sqrt{3} \mathrm{~m} \approx 750 \sqrt{3} \end{aligned} $

Given, $\theta_p=30^{\circ}$ (Horizontal)

$ \begin{aligned} \theta_Q & =30^{\circ}(\text { Vertical }) \\\\ & =90^{\circ}-30^{\circ}=60^{\circ} \end{aligned} $

thus, $\quad \theta_Q=60^{\circ}$ (Horizontal)

Maximum range reached,

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

For body $P$,

$ R_p=\frac{v_1^2 \sin 60^{\circ}}{g} \quad \ldots \text { (i) } $

For body $Q$

$ R_Q=\frac{v_2^2 \sin 120^{\circ}}{g} \quad \ldots \text { (ii) } $

The ratio of $\frac{R_P}{R_Q}=\frac{v_1^2}{v_2^2}=\frac{1}{2}$

$ \begin{aligned} \frac{H_P}{H_Q} & =\frac{v_1^2}{v_2^2} \times \frac{\sin ^2 30^{\circ}}{\sin ^2 60^{\circ}} \\\\ & =\frac{1}{2} \times \frac{1}{4} \times \frac{4}{3}=\frac{1}{6} \end{aligned} $

Given, Initial velocity, $u=\hat{\mathbf{i}}+2 \hat{\mathbf{j}} \mathrm{~m} / \mathrm{s}$

Acceleration due to gravity $=g=10 \mathrm{~m} / \mathrm{s}^2$

$ \begin{aligned} & \tan \theta=\frac{P}{B}=\frac{2}{1} \\ & \cos \theta=\frac{1}{\sqrt{5}} \end{aligned} $

$ \begin{aligned} &\text { Using the formula for path equation }\\ &\begin{aligned} Y & =x \tan \theta-\frac{g x^2}{2 u^2 \cos ^2 \theta} \\ |u| & =\sqrt{1^2+2^2}=\sqrt{5} \\ Y & =X \times 2-\frac{10 \times x^2}{2 \times 5 \times \frac{1}{5}} \\ y & =2 x-5 x^2 \end{aligned} \end{aligned} $

Given,

Horizontal Range, $R=80 \mathrm{~m}$

Angle $\theta=45^{\circ}$

As, the range of projectile is maximum when the angle of projection is $45^{\circ}$

Using the formula for range.

$ \begin{aligned} & R=\frac{u^2 \sin 2 \theta}{g} \\ & 80 \times 10=u^2 \quad\left[g=10 \mathrm{~m} / \mathrm{s}^2\right] \end{aligned} $

Now using $3^{\text {rd }}$ equation of motion for vertical motion,

$ v^2=u^2-2 a s $

(direction of $\mathbf{u}$ and $\mathbf{g}$ is opposite in nature)

at highest point $v=0$

$ \begin{aligned} u^2 & =2 g s \\ s & =\frac{u^2}{2 g} \end{aligned} $

( $s$ is height from horizontal level)

$ s=\frac{800}{2 \times 10}=40 \mathrm{~m} $

Given,

Magnitude of velocity, $v=10 \mathrm{~m} / \mathrm{s}$

Angle with $X$-axis $=60^{\circ}$

Using vector component resolution method,

The component of velocity along $X$-axis

$ =v \cos \theta \hat{\mathbf{i}} $

Component of velocity along $Y$-axis

$ \begin{aligned} & =v \sin \theta \hat{\mathbf{j}} \\ \mathbf{v} & =\mathbf{v}_x \hat{\mathbf{i}}+\mathbf{v}_y \hat{\mathbf{j}} \\ & =v \cos \theta \hat{\mathbf{i}}+v \sin \theta \hat{\mathbf{j}} \\ & =10 \times \cos 60 \hat{\mathbf{i}}+10 \times \sin 60 \hat{\mathbf{j}} \\ & =10 \times \frac{1}{2} \hat{\mathbf{i}}+10 \times \frac{\sqrt{3}}{2} \hat{\mathbf{j}} \\ \mathbf{v} & =5 \hat{\mathbf{i}}+5 \sqrt{3} \hat{\mathbf{j}} \end{aligned} $

Given, initial velocity $(u)=50 \mathrm{~ms}^{-1}$

angle of projection, $(\theta)=30^{\circ}$

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

$ =\frac{2 \times 50 \times \sin 30^{\circ}}{10}=\frac{100 \times\left(\frac{1}{2}\right)}{10}=5 \mathrm{~s} $

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

$ =\frac{2500 \times\left(\frac{\sqrt{3}}{2}\right)}{10}=216.5 \mathrm{~m} $

Now, we have time $t=3 \mathrm{~s}$

$ \frac{t}{T}=\frac{x}{R} $

$ \begin{array}{ll} \Rightarrow & \frac{3}{5}=\frac{x}{216.5} \\ \Rightarrow & x=\frac{3}{5} \times 216.5=129.9 \mathrm{~m} \end{array} $

The distance beyond the well where the stone will hit

$ R-x=216.5-129.9=86.6 m$

Given,

Each step distance is equal.

1st $\rightarrow 2$ step towards east.

2nd $\rightarrow$ take right and walks 4 steps south

3rd $\rightarrow$ take right and walks 6 steps towards west.

4th $\rightarrow$ take right and walks.

Total steps $=20$.

Draw the given data,

At last turn, person travels

$ =20-6+4+2=8 \text { steps } $

Thus, final direction is north west.

Given,

$ \begin{aligned} \text { Initial speed of plane } & =60 \mathrm{~km} / \mathrm{h} \\ & =\frac{50}{3} \mathrm{~m} / \mathrm{s} \end{aligned} $

Vertical height $\mp 490 \mathrm{~m}$Horizontal distance

…(i)

$ \begin{array}{ll} s=u \times t & \left(s=u t+\frac{1}{2} a t^2\right) \\ (a=0) & (s) \end{array} $

For vertical component of motion,

$ u=0 $

using equation of motion,

$ \begin{aligned} & s=u t+\frac{1}{2} a t^2(a=g) \\ & t=\sqrt{\frac{2 s}{g}} \Rightarrow t=\sqrt{\frac{2 \times 490}{9.8}} \\ & t=10 \mathrm{~s} \end{aligned} $

Put this in Eq. (i)

$ \begin{aligned} & s=\frac{50}{3} \times 10 \\ & s=\frac{500}{3} \mathrm{~m} \end{aligned} $

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} $

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} $

The given situation is shown below :

According to given condition, range must be greater than 20 m and less than 24.2 m .

$ \Rightarrow \quad 20 \leq R \leq 24.2 $

$ \begin{aligned} &\begin{aligned} & \Rightarrow \quad 20 \leq \frac{u^2 \sin 2 \theta}{g} \leq 24.2 \\ & \Rightarrow \quad 20 \leq \frac{u^2 \times \sin \left(2 \times 15^{\circ}\right)}{10} \leq 24.2 \\ & \Rightarrow \quad 200 \leq u^2 \times 1 / 2 \leq 242\left(\because \sin 30^{\circ}=\frac{1}{2}\right) \\ & \Rightarrow \quad 400 \leq u^2 \leq 484 \\ & \Rightarrow \quad 20 \mathrm{~m} / \mathrm{s} \leq u \leq 22 \mathrm{~m} / \mathrm{s} \end{aligned}\\ &\text { So, option (a) is best choice. } \end{aligned} $