Motion in a Plane

Let u be the initial velocity and $\theta=45^{\circ}$ be the angle of projection.

The initial vertical velocity is $\mathrm{u}_{\mathrm{y}}=\mathrm{u} \sin \theta$

The ball attains maximum height in $\mathrm{t}_{\max }=2 \mathrm{~s}$.

At the maximum height, the vertical component of the velocity becomes zero $\left(\mathrm{v}_{\mathrm{y}}=0\right)$.

Using the first equation of motion $(\mathrm{v}=\mathrm{u}+\mathrm{at})$ :

$ \mathrm{v}_{\mathrm{y}}=\mathrm{u}_{\mathrm{y}}-\mathrm{gt} $

$\Rightarrow $ $ 0=\mathrm{u}_{\mathrm{y}}-(10)(2) \Rightarrow \mathrm{u}_{\mathrm{y}}=20 \mathrm{~m} / \mathrm{s} $

So, the initial vertical velocity is $20 \mathrm{~m} / \mathrm{s}$.

The ball lands on the roof of the building at $\mathrm{t}=3 \mathrm{~s}$.

This means at $t=3$, the vertical displacement $(y)$ of the ball is equal to the height of the building $(H)$.

Using the second equation of motion $\left(s=u t+\frac{1}{2} a t^2\right)$ :

$ \mathrm{y}=\mathrm{u}_{\mathrm{y}} \mathrm{t}-\frac{1}{2} \mathrm{gt}^2 $

Substituting the given values :

• 1. $y=H$ (Height of the building)

• 2. $\mathrm{u}_{\mathrm{y}}=20 \mathrm{~m} / \mathrm{s}$ (Calculated in Step 2)

• 3. $\mathrm{t}=3 \mathrm{~s}$ (Given time of landing)

$ H=(20)(3)-\frac{1}{2}(10)(3)^2=15 \mathrm{~m} $

Therefore, the height of the building is 15 m. Hence, the correct option is (D).

When a projectile is launched into the air, it moves in two dimensions: horizontally ( $\mathrm{x}-$ axis ) and vertically ( y - axis).

Along vertical direction gravity constantly pulls the projectile downward. Therefore, there is a constant downward acceleration $\left(\mathrm{a}_{\mathrm{y}}=-\mathrm{g}\right)$.

Along horizontal direction there are no horizontal forces (assuming no air resistance). According to Newton's First Law, an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force. Therefore, the horizontal acceleration is zero ( $\mathrm{a}_{\mathrm{x}}=0$ ).

Because $\mathrm{a}_{\mathrm{x}}=0$, the horizontal component of the projectile's velocity remains perfectly constant throughout its entire journey.

If a projectile is launched with an initial speed u at an angle $\theta$, its horizontal component is

$ \mathrm{u}_{\mathrm{x}}=\mathrm{u} \cos \theta $

At any later point in time, if the projectile has a new speed v and is moving at a new angle $\alpha$, its horizontal component is :

$ \mathrm{v}_{\mathrm{x}}=\mathrm{v} \cos \alpha $

Since the horizontal velocity never changes, these two expressions must be equal :

$ \mathrm{u}_{\mathrm{x}}=\mathrm{v}_{\mathrm{x}} $

$\Rightarrow $ $ \mathrm{u} \cos \theta=\mathrm{v} \cos \alpha $

Substituting the given values :

$ \mathrm{u} \cos \left(60^{\circ}\right)=20 \cos \left(45^{\circ}\right) $

$\Rightarrow $ $u\left(\frac{1}{2}\right)=20 \times\left(\frac{1}{\sqrt{2}}\right)$

$\Rightarrow $ $u=20 \times \frac{2}{\sqrt{2}}$

Therefore, the initial speed of projectile is $20 \sqrt{2} \mathrm{~m} / \mathrm{s}$.

Hence, the correct option is (B).

The speed of river is $v_r=18 \times \frac{5}{18}=5 \mathrm{~m} / \mathrm{s}$

The speed of boat in still water is $\mathrm{v}_{\mathrm{b}}=36 \times \frac{5}{18}=10 \mathrm{~m} / \mathrm{s}$

The width of river is $\mathrm{w}=200 \mathrm{~m}$

The time taken to cross a river is determined solely by the component of the boat's velocity perpendicular to the river banks.

$ t=\frac{w}{v_b \sin \theta} $

Where theta is the angle with the bank. Time is minimum when the boat heads straight across ( $\theta=90^{\circ}$ ), so $\sin \theta=1$.

$ t_{\min 1 \rightarrow 2}=\frac{w}{v_b}=\frac{200}{10}=20 \mathrm{~s} $

Also, while returning to initial bank in minimum time,

$ t_{\min 2 \rightarrow 1}=\frac{w}{v_b}=\frac{200}{10}=20 \mathrm{~s} $

Total time of journey is,

$ \mathrm{T}_{\text {total }}=\mathrm{t}_{\min 1 \rightarrow 2}+\mathrm{t}_{\min 2 \rightarrow 1}=2 \times 20=40 \mathrm{~s} $

When the boat heads straight across to minimize time, the river flow carries it downstream. This horizontal distance is,

Drift (one way) $=\mathrm{v}_{\mathrm{r}} \times \mathrm{t}_{\text {min } 1 \rightarrow 2}=5 \times 20 \mathrm{~s}=100 \mathrm{~m}$ east

The boat again heads straight across to minimize time. The river still flows from west to east, so the boat drifts another 100 m east.

Total Displacement along the bank:

$ \Delta \mathrm{x}=100 \mathrm{~m} \text { (east) }+100 \mathrm{~m} \text { (east) }=200 \mathrm{~m} $

Therefore, the minimum time is 40 s and the displacement along the bank is 200 m . Hence, the correct option is (C).

For two projectiles launched with the same initial speed u to have the same horizontal range R, their angles of projection must be complementary.

Let the first angle be $\theta$. Then the second angle must be ( $90^{\circ}-\theta$ ) .

The time taken for the projectile to return to the same horizontal level.

$\mathrm{T}=\frac{2 \mathrm{u} \sin \theta}{\mathrm{~g}} \ldots (i) $

And the total horizontal distance covered is the range of projectile, R .

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

$\Rightarrow $ $\mathrm{R}=\frac{2 \mathrm{u}^2 \sin \theta \cos \theta}{\mathrm{~g}}........ (ii) $

For the first body $\left(\mathrm{T}_1=5 \mathrm{~s}\right)$ :

$ \mathrm{T}_1=\frac{2 \mathrm{u} \sin \theta}{\mathrm{~g}} \Rightarrow \sin \theta=\frac{\mathrm{g} \mathrm{~T}_1}{2 \mathrm{u}} $

For the second body ( $\mathrm{T}_2=10 \mathrm{~s}$ ):

$ \mathrm{T}_2=\frac{2 \mathrm{u} \sin \left(90^{\circ}-\theta\right)}{\mathrm{g}} \Rightarrow \cos \theta=\frac{\mathrm{g} \mathrm{~T}_2}{2 \mathrm{u}} $

$\Rightarrow \mathrm{T}_1 \times \mathrm{T}_2=\left(\frac{2 \mathrm{u} \sin \theta}{\mathrm{g}}\right) \times\left(\frac{2 \mathrm{u} \cos \theta}{\mathrm{g}}\right)$

$\Rightarrow $ $\mathrm{T}_1 \mathrm{~T}_2=\frac{2}{\mathrm{~g}}\left(\frac{2 \mathrm{u}^2 \sin \theta \cos \theta}{\mathrm{~g}}\right)$

Using the relation (ii)

$ \mathrm{T}_1 \mathrm{~T}_2=\frac{2 \mathrm{R}}{\mathrm{~g}} $

$\Rightarrow $ $ \mathrm{R}=\frac{\mathrm{g} \mathrm{~T}_1 \mathrm{~T}_2}{2} $

Here, the time of flight of first projectile is $\mathrm{T}_1=5 \mathrm{~s}$ and that of second projectile is it $\mathrm{T}_2=10 \mathrm{~s}$ Putting the values in the formula of horizontal range,

$ \mathrm{R}=\frac{10 \times 5 \times 10}{2} $

$\Rightarrow $ $ \mathrm{R}=250 \mathrm{~m} $

Therefore, the value of the horizontal range R is 250 m.

Hence, the correct option is (A).

Mass of the body is

$ m=100\text{ g}=0.1\text{ kg} $

The force acting on it is

$ \vec F = 5\hat i + 10\hat j $

Using Newton’s second law,

$ \vec a=\frac{\vec F}{m} $

So,

$ \vec a=\frac{5\hat i+10\hat j}{0.1} =50\hat i+100\hat j $

Hence,

• acceleration in $x$-direction: $a_x=50\text{ m/s}^2$

• acceleration in $y$-direction: $a_y=100\text{ m/s}^2$

Since the body starts moving at $t=0$, we take initial velocity as zero.

Using

$ s=ut+\frac{1}{2}at^2 $

for $t=2\text{ s}$:

In $x$-direction,

$ s_x=0+\frac{1}{2}(50)(2)^2 =25\times 4=100\text{ m} $

In $y$-direction,

$ s_y=0+\frac{1}{2}(100)(2)^2 =50\times 4=200\text{ m} $

Given position after $2$ s is

$ (2x\hat i+5y\hat j)\text{ m} $

So,

$ 2x=100 \quad\Rightarrow\quad x=50 $

and

$ 5y=200 \quad\Rightarrow\quad y=40 $

Therefore,

$ x:y=50:40=5:4 $

So the correct answer is:

$ \boxed{5:4} $

Option D

For a projectile, the angle it makes with the horizontal at any instant is found from the velocity components.

Given:

$ x=24t $

and

$ y=43.6t-4.9t^2 $

So, horizontal velocity is

$ v_x=\frac{dx}{dt}=24 $

and vertical velocity is

$ v_y=\frac{dy}{dt}=43.6-9.8t $

At $t=2\text{ s}$,

$ v_y=43.6-9.8(2)=43.6-19.6=24 $

Thus,

$ \tan \theta=\frac{v_y}{v_x}=\frac{24}{24}=1 $

So,

$ \theta=45^\circ $

Hence, the correct answer is:

$ \boxed{45^\circ} $

Option B

Given that the maximum height reached by the first ball is 8 times that of the second ball:

$ \left(H_{\max}\right)_1 = 8 \times \left(H_{\max}\right)_2 $

We can relate this to the initial velocities and angles using the formula for maximum height:

$ \frac{u^2 \sin^2 \theta_1}{2g} = 8 \times \frac{u^2 \sin^2 \theta_2}{2g} $

Simplifying, we find:

$ \sin \theta_1 = 2 \sqrt{2} \sin \theta_2 $

Now, the ratio of the total flying times $T_1$ and $T_2$ is given by:

$ \frac{T_1}{T_2} = \frac{2u \sin \theta_1 / g}{2u \sin \theta_2 / g} = \frac{\sin \theta_1}{\sin \theta_2} = 2 \sqrt{2} $

Thus, the ratio of $T_1$ to $T_2$ is $2 \sqrt{2} : 1$.

$\begin{aligned} & u=360 \times \frac{5}{18}=100 \mathrm{~m} / \mathrm{s} \\ & \mathrm{x}=\mathrm{u} \times \mathrm{t}=2 \times 10^3 \mathrm{~m} \\ & \mathrm{t}=\sqrt{\frac{2 \mathrm{H}}{\mathrm{~g}}} \Rightarrow \mathrm{H}=\frac{\mathrm{t}^2 \mathrm{~g}}{2} \\ & \mathrm{H}=\frac{400 \times 10}{2} \\ & \mathrm{H}=2000 \mathrm{~m} \\ & \mathrm{D}=\sqrt{\mathrm{x}^2+\mathrm{H}^2} \\ & \mathrm{D}=2 \sqrt{2} \mathrm{~km} \end{aligned}$

To find the ratio of the times of flight for two projectiles fired from the ground with the same initial speed but at different angles, we follow these steps:

Projectile Angles:

The first projectile is launched at an angle $\theta_1 = 45^\circ + \alpha$.

The second projectile is launched at an angle $\theta_2 = 45^\circ - \alpha$.

Time of Flight Formula:

The time of flight $T$ for a projectile is given by:

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

where $v$ is the initial speed and $g$ is the acceleration due to gravity.

Ratio of Times of Flight:

The ratio of the times of flights $\frac{T_1}{T_2}$ can be determined as follows:

$ \frac{T_1}{T_2} = \frac{\sin(45^\circ + \alpha)}{\sin(45^\circ - \alpha)} $

Applying Trigonometric Identities:

Use the sine addition and subtraction formulas:

$ \sin(45^\circ + \alpha) = \frac{1}{\sqrt{2}}(\cos \alpha + \sin \alpha) $

$ \sin(45^\circ - \alpha) = \frac{1}{\sqrt{2}}(\cos \alpha - \sin \alpha) $

Simplifying the Ratio:

Substitute the expressions for $\sin(45^\circ + \alpha)$ and $\sin(45^\circ - \alpha)$ into the ratio:

$ \frac{T_1}{T_2} = \frac{\frac{1}{\sqrt{2}} (\cos \alpha + \sin \alpha)}{\frac{1}{\sqrt{2}} (\cos \alpha - \sin \alpha)} = \frac{\cos \alpha + \sin \alpha}{\cos \alpha - \sin \alpha} $

Factor out the trigonometric functions to express in terms of tangent:

$ \frac{T_1}{T_2} = \frac{1 + \tan \alpha}{1 - \tan \alpha} $

Therefore, the ratio of the times of flights for the two projectiles is $\frac{1 + \tan \alpha}{1 - \tan \alpha}$.

Given that the horizontal range of a projectile is three times its maximum height, we need to determine the value of $ n $ in the expression for the range $\frac{n u^2}{25 g}$.

Let's start by setting up the equations for the range and maximum height of a projectile.

The formula for the range $ R $ is:

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

The formula for the maximum height $ H_{\max} $ is:

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

According to the problem, the range $ R $ is three times the maximum height $ H_{\max} $:

$ R = 3H_{\max} $

Substituting the formulas for $ R $ and $ H_{\max} $ into this condition gives:

$ \frac{u^2 \sin 2\theta}{g} = 3 \cdot \frac{u^2 \sin^2 \theta}{2g} $

Simplifying this equation, we get:

$ 2 \sin \theta \cos \theta = \frac{3}{2} \sin^2 \theta $

Using the identity $ \sin 2\theta = 2 \sin \theta \cos \theta $, we can rewrite:

$ 2 \sin \theta \cos \theta = \sin 2\theta $

This gives us:

$ \sin 2\theta = \frac{3}{2} \sin^2 \theta $

Dividing both sides by $\sin \theta$, assuming $\sin \theta \neq 0$, we have:

$ 2 \cos \theta = \frac{3}{2} \sin \theta $

Rearranging the terms to solve for $\tan \theta$, we get:

$ \tan \theta = \frac{4}{3} $

Thus, the angle $\theta$ where $\tan \theta = \frac{4}{3}$ corresponds to $\theta = 53^\circ$.

Now, substituting back into the range formula:

$ R = \frac{u^2 \cdot 2 \cdot \frac{3}{5} \cdot \frac{4}{5}}{g} $

The calculation yields:

$ R = \frac{24 u^2}{25 g} $

Therefore, the value of $ n $ in the expression $\frac{n u^2}{25 g}$ is $ n = 24 $.

$\mathrm{H}_{\max }=\frac{\mathrm{u}^2 \cos ^2 \phi}{2 \mathrm{~g}}$

Step 1: Find the boat’s speed across the river

The boat's maximum speed is $27~\mathrm{km/h}$, and it makes an angle of $150^\circ$ with the direction of the river.

The part of the boat’s speed that helps it cross the river straight (perpendicular to the river) is found by: $27 \times \cos 60^\circ$.

Since $\cos 60^\circ = \frac{1}{2}$, this gives: $27 \times \frac{1}{2} = \frac{27}{2}~\mathrm{km/h}$.

Step 2: Change speed to meters per second

$\frac{27}{2}~\mathrm{km/h} = \frac{27}{2} \times \frac{1000}{3600} = \frac{27}{2} \times \frac{5}{18}~\mathrm{m/s}$.

Step 3: Use the time the boat takes

The boat takes $30$ seconds to cross.

Distance = speed × time
So the width of the river is:

$\mathrm{S} = \frac{27}{2} \times \frac{5}{18} \times 30~\mathrm{m}$

Step 4: Do the calculation

Calculate the value:
$\frac{27}{2} \times \frac{5}{18} \times 30 = 112.5~\mathrm{m}$

So, the width of the river is $112.5$ meters.

$\begin{aligned} & \mathrm{H}_{\mathrm{Max}}=\frac{(\mathrm{usin} \theta)^2}{2 \mathrm{~g}} \\ & \frac{\left(\mathrm{H}_{\max }\right)_1}{\left(\mathrm{H}_{\max }\right)_2}=\frac{\mathrm{u}^2 \sin ^2(45-\alpha)}{\mathrm{u}^2 \sin ^2(45+\alpha)} \\ & =\frac{\left(\frac{1}{\sqrt{2}} \cos \alpha-\frac{1}{\sqrt{2}} \sin \alpha\right)^2}{\left(\frac{1}{\sqrt{2}} \cos \alpha+\frac{1}{\sqrt{2}} \sin \alpha\right)^2} \\ & =\frac{1-\sin 2 \alpha}{1+\sin 2 \alpha} \end{aligned}$

$\begin{aligned} & \overrightarrow{\mathrm{r}}=5 \mathrm{t}^2 \hat{\mathrm{i}}-5 \hat{\mathrm{t}} \\ & \overrightarrow{\mathrm{v}}=10 \mathrm{t} \hat{\mathrm{i}}-5 \hat{\mathrm{j}} \\ & \overrightarrow{\mathrm{v}}=20 \hat{\mathrm{i}}-5 \hat{\mathrm{j}} \quad \text { at } \mathrm{t}=2 \mathrm{sec} \end{aligned}$

$\begin{aligned} & \tan \theta=\frac{20}{5}=4 \\ & \theta=\tan ^{-1} 4 \\ & \text { From-veY-axis } \end{aligned}$

$ \vec{v_0} = \frac{v_0}{\sqrt{2}}\,\hat{i} + \frac{v_0}{\sqrt{2}}\,\hat{j}. $

At maximum height, the vertical component of the velocity becomes zero. Setting the vertical velocity to zero:

$ \frac{v_0}{\sqrt{2}} - g\,t = 0 \quad \Longrightarrow \quad t = \frac{v_0}{g\sqrt{2}}. $

The coordinates at this time are:

$ x = \frac{v_0}{\sqrt{2}}\,t = \frac{v_0}{\sqrt{2}} \cdot \frac{v_0}{g\sqrt{2}} = \frac{v_0^2}{2g}, $

$ y = \frac{v_0}{\sqrt{2}}\,t - \frac{1}{2}g\,t^2 = \frac{v_0^2}{2g} - \frac{1}{2}g\,\frac{v_0^2}{2g^2} = \frac{v_0^2}{2g} - \frac{v_0^2}{4g} = \frac{v_0^2}{4g}. $

Thus, the position vector at maximum height is:

$ \vec{r} = \frac{v_0^2}{2g}\,\hat{i} + \frac{v_0^2}{4g}\,\hat{j}. $

At maximum height, the only nonzero component of velocity is the horizontal one:

$ \vec{v} = \frac{v_0}{\sqrt{2}}\,\hat{i}. $

The angular momentum about the origin is given by:

$ \vec{L} = m\,\vec{r} \times \vec{v}. $

Writing out the cross product:

$ \vec{r} \times \vec{v} = \left(\frac{v_0^2}{2g}\,\hat{i} + \frac{v_0^2}{4g}\,\hat{j}\right) \times \frac{v_0}{\sqrt{2}}\,\hat{i}. $

Since the cross product of $\hat{i}$ with itself is zero and:

$ \hat{j} \times \hat{i} = -\hat{k}, $

we have:

$ \vec{r} \times \vec{v} = \frac{v_0^2}{4g} \cdot \frac{v_0}{\sqrt{2}}\, (-\hat{k}) = -\frac{v_0^3}{4\sqrt{2}g}\,\hat{k}. $

Thus, the angular momentum is:

$ \vec{L} = -\frac{m\,v_0^3}{4\sqrt{2}g}\,\hat{k}. $

This means the magnitude is:

$ \frac{m\,v_0^3}{4\sqrt{2}g}, $

and its direction is along the negative $\hat{k}$ (or negative $z$-axis).

$\begin{aligned} & \mathrm{k}_{\mathrm{i}}=\frac{1}{2} \mathrm{mv}^2 \\ & \mathrm{k}_{\mathrm{f}}=\frac{1}{2} \mathrm{~m}\left(\mathrm{v} \cos 60^{\circ}\right)^2=\frac{1}{8} \mathrm{mv}^2 \\ & \Delta \mathrm{k}=\mathrm{k}_{\mathrm{i}}-\mathrm{k}_{\mathrm{f}}=\frac{3}{8} \mathrm{mv}^2=\frac{3}{8} \times 0.1 \times 400=15 \mathrm{~J} \end{aligned}$

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

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} $

$ \begin{aligned} & \text { } R_1=R_2 \text {, Thus, } \theta_1=90-\theta_2 \\ & \Rightarrow \quad \theta_2=90-\theta_1 \end{aligned} $

$ \begin{aligned} \therefore \quad \frac{T_1}{T_2} & =\frac{\frac{2 u \sin \theta_1}{g}}{\frac{2 u \sin \theta_2}{g}}=\frac{\sin \theta_1}{\sin \theta_2} \\ & =\frac{\sin \theta_1}{\sin \left(90-\theta_1\right)}=\frac{\sin \theta_1}{\cos \theta_1} \\ & =\tan \theta_1 \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } R=3 H \\ & \quad \frac{u^2 \sin 2 \theta}{g}=\frac{3 u^2 \sin ^2 \theta}{2 g} \\ & \Rightarrow \quad 2 \sin \theta \cos \theta=\frac{3}{2} \sin ^2 \theta \\ & \Rightarrow \quad \tan \theta=\frac{4}{3} \\ & \therefore \quad \sin \theta=\frac{4}{5} \text { and } \cos \theta=\frac{3}{5} \end{aligned} \end{aligned} $

$ \begin{aligned} \therefore \quad & R=\frac{u^2 \sin 2 \theta}{g}=\frac{u^2}{g} \times 2 \sin \theta \cos \theta \\ & =\frac{u^2}{g} \times 2 \times \frac{4}{5} \times \frac{3}{5} \\ & =\frac{24 u^2}{25 g} \end{aligned} $

$ \begin{aligned} &\text { At maximum height }\\ &\begin{aligned} \Rightarrow \quad & u \cos \theta=20 \mathrm{~m} / \mathrm{s} \\ \Rightarrow \quad & u \cos 45^{\circ}=20 \\ \Rightarrow \quad & u=20 \sqrt{2} \mathrm{~m} / \mathrm{s} \\ \therefore \quad & H=\frac{u^2 \sin ^2 \theta}{2 g}=\frac{(20 \sqrt{2})^2 \sin ^2 45^{\circ}}{2 \times 10} \\ & =20 \mathrm{~m} \end{aligned} \end{aligned} $

Acceleration produced in the body due to applied force

$ a=\frac{F}{m}=\frac{3}{2}=1.5 \mathrm{~m} / \mathrm{s}^2 $

Since, force is applied normal to the direction of its initial velocity, thus, $v_y=0+a t=1.5 \times 2=3 \mathrm{~m} / \mathrm{s}$

∴ Initial velocity, $v_x=4 \mathrm{~m} / \mathrm{s}$

∴ Resultant velocity, $v=\sqrt{v_x^2+v_y^2}$

$ =\sqrt{4^2+3^2}=5 \mathrm{~m} / \mathrm{s} $

Step 1: Write the formula for range

The formula to find the range $R$ when a body is projected is:

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

Step 2: Put in the given values

Here, $R = 180 \sqrt{3}$ m, $u = 60$ m/s, and $g = 10$ m/s$^2$.

Step 3: Arrange the formula to find $\sin 2\theta$

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

Step 4: Substitute the values

$ \sin 2\theta = \frac{180 \sqrt{3} \times 10}{60 \times 60} $

$ \sin 2\theta = \frac{1800 \sqrt{3}}{3600} = \frac{\sqrt{3}}{2} $

Step 5: Identify the angle for $\sin 2\theta$

We know that $\sin 60^\circ = \frac{\sqrt{3}}{2}$.

So, $2\theta = 60^\circ$

$\theta = 30^\circ$

Step 6: Check for other possible angles

The range at angle $\theta$ is the same as at $90^\circ - \theta$.

So, the angle of projection can be $30^\circ$ or $60^\circ$.

Considering vertical motion,

$ \begin{aligned} & u_y=u \sin \theta, S_y=60 \mathrm{~m} \\ & a_y=-10 \mathrm{~ms}^{-2}, t=2 \mathrm{~s} \end{aligned} $

Using $S_y=u_y t+\frac{1}{2} a_y t^2$

we get,

$ \begin{aligned} & 60=u \sin \theta \times 2-\frac{1}{2} \times 10 \times 2^2 \\ \Rightarrow & u \sin \theta=40 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Now, time of flight is

$ T=\frac{2 n \sin \theta}{g}=\frac{2 \times 40}{10}=8 \mathrm{~s} $

Range of projectile $R=40 \mathrm{~m}$

$ \begin{aligned} & R=40=\frac{u^2 \sin 2 \theta}{g} \\ \Rightarrow & u^2=\frac{40 g}{\sin 2 \theta} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

and from equation of trajectory

$ \begin{aligned} & Y=x \tan \theta-\frac{1}{2} g \frac{x^2}{u^2 \cos ^2 \theta} \\ & Y=x \tan \theta-\frac{1}{2} \frac{g x^2 \sin 2 \theta}{40 g \times \cos ^2 \theta} \\ & 20=30 \tan \theta-\frac{1}{80} \times \frac{900 \times 2 \sin \theta \cos \theta}{\cos ^2 \theta} \\ & 20=30 \tan \theta-\frac{1800}{80} \tan \theta \\ & 20=30 \tan \theta-\frac{45}{2} \tan \theta \\ & 20=\frac{15 \tan \theta}{2} \\ & \Rightarrow 15 \tan \theta=40 \\ & \quad \tan \theta=\frac{40}{15}=\frac{8}{3} \end{aligned} $

$ \theta=\tan ^{-1}\left(\frac{8}{3}\right) $

$y=A x-B x^2$

Comparing with equation of trajectory,

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

we get,

$ \begin{aligned} & A=\tan \theta \text { and } B=\frac{g}{2 u^2 \cos ^2 \theta} \\ & \therefore \quad \frac{H}{R}=\frac{\frac{u^2 \sin ^2 \theta}{2 g}}{\frac{u^2 \sin 2 \theta}{g}}=\frac{\sin ^2 \theta}{2 \sin 2 \theta} \\ & \quad=\frac{\sin ^2 \theta}{4 \sin \theta \cos \theta}=\frac{1}{4} \tan \theta=\frac{A}{4} \end{aligned} $

Height of auditorium is 30 m and ball just reaches to this height so this is the maximum height of projectile.

Let angle of projection is $\theta$.

$ \begin{aligned} & \text { So, } H_{\max }=30=\frac{u^2 \sin ^2 \theta}{2 g} \\ & \qquad \begin{aligned} 30 & =\frac{(30)^2 \sin ^2 \theta}{2 \times 10} \\ \sin ^2 \theta & =\frac{600}{900}=\frac{2}{3} \end{aligned} \\ & \Rightarrow \sin \theta=\sqrt{\frac{2}{3}} \\ & \text { and } \cos \theta=\frac{1}{\sqrt{3}} \end{aligned} $

Now, length of auditorium will be equal to range of projectile.

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

$ \begin{aligned} & =\frac{u^2(2 \sin \theta \cos \theta)}{g} \\ R & =\frac{(30)^2 \times 2 \times \sqrt{\frac{2}{3}} \times \frac{1}{\sqrt{3}}}{0.10} \\ R & =\frac{1800 \sqrt{2}}{30} \\ & =60 \sqrt{2} \mathrm{~m} \end{aligned} $

$y=b x^2$

$ \begin{aligned} \therefore \quad v_y & =\frac{d y}{d t}=\frac{d}{d t} \cdot b x^2=z b x \frac{d x}{d y} \\ v_y & =z b x v_x \end{aligned} $

differentiating again

$ \begin{aligned} & & \frac{d v_y}{d t} & =2 b\left(x \frac{d v_x}{d t}+v_x \frac{d x}{d t}\right) \\ & \Rightarrow & a_y & =2 b\left[x \cdot a_x+v_x \cdot v_x\right] \\ & \Rightarrow & a_y & =2 b\left[x a_x+v_x^2\right] \end{aligned} $

Since, $a_x=0$ and $a_y=a$

$ \begin{array}{ll} \therefore & a=2 b\left(0+v_x^2\right) \Rightarrow a=2 b v_x^2 \\ \Rightarrow & v_x=\sqrt{\frac{a}{2 b}} \end{array} $

If $v_1$ be the velocity at the time of projection and $v_2$ be the velocity at height 25 m , then

$ v_2^2=v_1^2-2 g \times 25 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

If $t$ be the time taken to reach at maximum height $H$ from point $B$, then

$ \begin{aligned} & 0=v_2-g t \\ & v_2=g t=10 \times 2=20 \mathrm{~m} / \mathrm{s} \end{aligned} $

$\therefore$ From eq. (i),

$ \begin{aligned} & 20^2=v_1^2-2 \times 10 \times 25 \\ \Rightarrow \quad & 400+500=v_1^2 \\ \Rightarrow \quad & v_1=\sqrt{900}=30 \mathrm{~m} / \mathrm{s} \end{aligned} $

Initial velocity, $\mathbf{u}=4 \hat{\mathbf{i}}$

$ \begin{aligned}Final \,\,velocity, & \mathbf{v}=4 \sqrt{2}\left(\frac{1}{\sqrt{2}}(\hat{\mathbf{i}}+\hat{\mathbf{j}})\right) \\ & =4 \hat{\mathbf{i}}+4 \hat{\mathbf{j}} \end{aligned} $

$ \begin{aligned}Displacement\,\, \mathbf{r} & =\mathbf{v}_{\text {avg }} \times t \\ \mathbf{r} & =\left(\frac{\mathbf{v}+\mathbf{u}}{2}\right) \times t \\ & =\left(\frac{4 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+4 \hat{\mathbf{i}}}{2}\right) \times 4 \\ & =4(4 \hat{\mathbf{i}}+2 \hat{\mathbf{j}})=16 \hat{\mathbf{i}}+8 \hat{\mathbf{j}} \end{aligned} $

Average velocity

$ \begin{gathered} \mathbf{v}_{\text {avg }}=\frac{\text { Displacement covered }}{\text { Time taken }} \\ =\frac{16 \hat{\mathbf{i}}+8 \hat{\mathbf{j}}}{4}=4 \hat{\mathbf{i}}+2 \hat{\mathbf{j}} \\ \left(\mathbf{v}_{\text {avg }}\right)=\text { magnitude of average velocity } \\ =\sqrt{4^2+2^2} \\ =\sqrt{20}=2 \sqrt{5} \mathrm{~m} / \mathrm{s} \end{gathered} $

$u=30 \hat{\mathbf{i}}+40 \hat{\mathbf{j}}$

$ \begin{aligned} & F=(-\hat{\mathbf{i}}+5 \hat{\mathbf{j}}) \mathrm{N} \\ & a=\frac{F}{m}=\left(\frac{-\hat{\mathbf{i}}+5 \hat{\mathbf{j}}}{5}\right)=-\left(\frac{\hat{\mathbf{i}}}{5}+\hat{\mathbf{j}}\right) \\ & u_x=40 \mathrm{~m} / \mathrm{s}, u_y=40 \mathrm{~m} / \mathrm{s} \end{aligned} $

$ \begin{aligned} &\text { In } y \text {-direction, }\\ &\begin{array}{ll} \therefore & a_y=-1 \mathrm{~m} / \mathrm{s}^2 \\ \therefore & v_y=u_y+a_y t \\ & 0=40-1 t \\ & t=40 \mathrm{~s} \end{array} \end{aligned} $

$ \begin{aligned} &\text { } A=\pi R^2=\pi\left(\frac{u^2 \sin 2 \theta}{g}\right)^2\\ &\begin{aligned} & =\pi \cdot\left[\frac{10^2 \sin \left(2 \times 45^{\circ}\right)}{10}\right]^2 \\ & =100 \pi \\ & =100 \times 3.14 \\ & =314 \mathrm{~m}^2 \end{aligned} \end{aligned} $

Let person catches ball after time ' $f$ '.

Then, $V_0 \cos \theta \cdot t=0.5 V_0 t$

$ \begin{aligned} & \Rightarrow \cos \theta=0.5=\frac{1}{2} \\ & \Rightarrow \cos \theta=\cos 60^{\circ} \\ & \Rightarrow \theta=60^{\circ} \end{aligned} $

To find the angle of projection for a projectile to have the same horizontal range and maximum height, we need to express both the range and the maximum height in terms of the projectile's initial velocity and the angle of projection, and then set them equal to each other.

The formula for the horizontal range $R$ of a projectile is given by:

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

where $v$ is the initial velocity of the projectile, $g$ is the acceleration due to gravity, and $\theta$ is the angle of projection.

The formula for the maximum height $H$ reached by the projectile is:

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

To have the same numerical value for $R$ and $H$, we set them equal to each other:

$\frac{v^2}{g} \sin 2\theta = \frac{v^2}{2g} \sin^2\theta.$

Simplifying this equation, we get:

$2 \sin 2\theta = \sin^2\theta.$

Using the double-angle formula, $\sin 2\theta = 2 \sin\theta \cos\theta$, we can rewrite the equation as:

$4 \sin\theta \cos\theta = \sin^2\theta.$

This can be simplified further to:

$4 \sin\theta \cos\theta = (\sin\theta)^2.$

Dividing both sides by $\sin\theta$ (assuming $\sin\theta \neq 0$), we get:

$4 \cos\theta = \sin\theta.$

Now, dividing both sides by $\cos\theta$, we have:

$4 = \tan\theta.$

So, the angle of projection $\theta$ is:

$\theta = \tan^{-1}(4).$

Therefore, the correct answer is:

Option D $\tan ^{-1}(4)$.

To determine the nature of the motion of the particle given by its coordinates in the $x$-$y$ plane, we analyze the given equations for $x$ and $y$ in terms of time $t$:

• $x = 2 + 4t$
• $y = 3t + 8t^2$

Firstly, the equation for $x$ is of the form $x = x_0 + vt$, where $x_0 = 2$ is the initial position and $v = 4$ is the constant velocity along the $x$-axis. This suggests a uniform motion along the $x$-axis because the velocity remains constant with time.

Secondly, the equation for $y$ is a second-degree polynomial in $t$, which indicates a parabolic path. The presence of the $t^2$ term ($8t^2$) signifies acceleration since the position along the $y$-axis is changing at a rate that itself changes over time.

The equation for $y$ can show two types of motion depending on the terms:

1. If it was of the form $y = y_0 + vt$, it would indicate uniform motion.
2. If it was of the form $y = y_0 + vt + \frac{1}{2}at^2$, where $a$ would represent acceleration, it would indicate uniformly accelerated motion. The presence of the $8t^2$ term here plays a similar role, indicating that the motion is uniformly accelerated in the $y$-direction due to the constant acceleration implied by this term.

Since the motion in the $y$ direction is determined by a quadratic equation, and the path of the particle depends on both the $x$ and $y$ coordinates, the motion of the particle is not along a straight line but rather follows a parabolic path due to the quadratic (second-degree) dependence on time in the $y$-coordinate.

Additionally, the acceleration is not changing with time, as deduced from the constant coefficient of the $t^2$ term in the $y$ equation, indicating uniform acceleration. Therefore, the motion is uniformly accelerated and follows a parabolic path.

Hence, the correct answer is:

Option D: uniformly accelerated having motion along a parabolic path.

For $\mathrm{u}_{\mathrm{A}}$ & $\mathrm{u}_{\mathrm{B}}$ time of flight and range can not be same. So above options are incorrect.

The position of an ant, denoted as $ \mathrm{S} $ in meters, moving in the $ \mathrm{Y} $-$ \mathrm{Z} $ plane is given by $ S = 2t^2 \hat{j} + 5 \hat{k} $, where $ t $ is in seconds. To determine the magnitude and direction of the ant's velocity at $ t = 1 $ second, we need to differentiate the position function with respect to time.

The velocity $ \overrightarrow{\mathrm{v}} $ is given by:

$\overrightarrow{\mathrm{v}} = \frac{d\mathrm{S}}{dt} = \frac{d}{dt} (2t^2 \hat{j} + 5 \hat{k})$

On differentiating, we get:

$\overrightarrow{\mathrm{v}} = 4t \hat{j}$

At $ t = 1 $ second:

$\overrightarrow{\mathrm{v}} = 4 \cdot 1 \hat{j} = 4 \hat{j}$

Therefore, the magnitude of the velocity is $ 4 \mathrm{~m/s} $ and it is directed along the $ y $-axis.

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} $

The path of the projectile is described by the equation:

$ Y = Px - Qx^2 $

Comparing this with the traditional equation of projectile motion:

$ Y = x \tan \theta - \frac{1}{2} g \frac{x^2}{u^2 \cos^2 \theta} $

we identify that:

$ P = \tan \theta \quad \text{(i)} $

$ Q = \frac{g}{2 u^2 \cos^2 \theta} \quad \text{(ii)} $

Using these relationships, we can derive the following:

Range $R$:

$ \frac{P}{Q} = \frac{\tan \theta}{g} = \frac{2u^2 \cos^2 \theta \cdot \tan \theta}{g} = \frac{u^2 \sin 2\theta}{g} = R $

Thus, $a \rightarrow \text{(i)}$.

Maximum Height ($H$):

$ \frac{P^2}{4Q} = \frac{\tan^2 \theta}{4 \cdot \frac{g}{2 u^2 \cos^2 \theta}} = \frac{\tan^2 \theta \cdot 2u^2 \cos^2 \theta}{4g} = \frac{u^2 \sin^2 \theta}{2g} = H $

Therefore, $b \rightarrow \text{(iii)}$.

Tangent of Projection Angle:

$ P = \tan \theta $

Hence, $d \rightarrow \text{(ii)}$.

Time of Flight ($T$):

$ \left(\sqrt{\frac{2}{gQ}}\right) P = \left(\sqrt{\frac{2}{g \cdot \frac{g}{2 u^2 \cos^2 \theta}}}\right) \tan \theta = \frac{2u \cos \theta}{g} \cdot \frac{\sin \theta}{\cos \theta} = \frac{2u \sin \theta}{g} = T $

Therefore, $c \rightarrow \text{(iv)}$.

This analysis establishes a clear correspondence between the projectile motion parameters and the terms given in the equation.

To determine the minimum height $ h $ from which the bowling machine releases the balls, consider the scenario where the ball is thrown horizontally. The landing velocity should make the smallest possible angle, which is $ 30^\circ $, with the horizontal.

Initial Conditions: At $ t = 0 $:

Horizontal velocity component: $ u_x = 10\sqrt{3} \, \text{ms}^{-1} $

Vertical velocity component: $ u_y = 0 \, \text{ms}^{-1} $

At time $ t = t $:

Horizontal velocity remains constant: $ v_x = 10\sqrt{3} $

Vertical velocity due to gravity: $ v_y = g t = 10t $

Condition for Minimum Angle:

The tangent of the angle of the landing velocity with respect to the horizontal is given by:

$ \tan 30^\circ = \frac{v_y}{v_x} $

Substituting the given values:

$ \frac{1}{\sqrt{3}} = \frac{10t}{10\sqrt{3}} $

Solve for $ t $:

$ \frac{1}{\sqrt{3}} = \frac{t}{\sqrt{3}} \Rightarrow t = 1 \, \text{s} $

Calculate Height $ h $:

Using the formula for vertical displacement under gravity:

$ h = \frac{1}{2} g t^2 = \frac{1}{2} \times 10 \times 1^2 = 5 \, \text{m} $

Thus, the height $ h $ from which the balls should be released so that the landing velocity makes the minimum angle of $ 30^\circ $ with the horizontal is $ 5 \, \text{meters} $.