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).

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

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.

Let the initial speed of both projectiles be v. The maximum height attained by a projectile can be calculated using the formula:

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

where H is the maximum height, v is the initial speed, θ is the angle of projection, and g is the acceleration due to gravity.

For the projectile projected at 30°, the maximum height is:

$H_1 = \frac{v^2 \sin^2 30^{\circ}}{2g} = \frac{v^2 \times \frac{1}{4}}{2g} = \frac{v^2}{8g}$

For the projectile projected at 60°, the maximum height is:

$H_2 = \frac{v^2 \sin^2 60^{\circ}}{2g} = \frac{v^2 \times \frac{3}{4}}{2g} = \frac{3v^2}{8g}$

Now, let's find the ratio of the maximum heights:

$\frac{H_1}{H_2} = \frac{\frac{v^2}{8g}}{\frac{3v^2}{8g}} = \frac{v^2}{3v^2}$

The v² terms cancel out, and we get:

$\frac{H_1}{H_2} = \frac{1}{3}$

Therefore, the ratio of the maximum heights attained by the two projectiles is 1 : 3

The range $R$ of a projectile launched with an initial speed $v$ and at an angle $\theta$ to the horizontal is given by:

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

where $g$ is the acceleration due to gravity.

From this equation, we can see that the range is dependent on the sine of twice the launch angle.

Given that the range at $15^\circ$ is $50$ m, if we launch the projectile at $45^\circ$ with the same velocity, we can compare the ranges by comparing $\sin(2 \times 15^\circ)$ and $\sin(2 \times 45^\circ)$:

$\sin(30^\circ) = \frac{1}{2}$

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

Therefore, the range at $45^\circ$ will be twice the range at $15^\circ$, because $\sin(90^\circ)$ is twice as large as $\sin(30^\circ)$.

So, the range when the projectile is launched at $45^\circ$ will be $2 \times 50$ m = $100$ m.

So, $100$ m is the correct answer.

The equation of the trajectory given is $y = x - \frac{x^2}{20}$.

This is a parabola, and it represents the path of the projectile.

The maximum height of the projectile corresponds to the vertex of the parabola.

The x-coordinate of the vertex for a parabola given by $y = ax^2 + bx + c$ is $-b/2a$.

In this case, $a = -1/20$ and $b = 1$, so the x-coordinate of the vertex is:

$ x_{\text{vertex}} = -\frac{b}{2a} = -\frac{1}{2 \times (-1/20)} = 10 $

Substituting this into the equation of the trajectory gives the y-coordinate of the vertex, which is the maximum height:

$ y_{\text{max}} = 10 - \frac{10^2}{20} = 10 - 5 = 5 $

So, the maximum height attained by the projectile is 5 m.

The range of a projectile launched with an initial velocity $v$ at an angle $\theta$ with respect to the horizontal is given by:

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

where $g$ is the acceleration due to gravity.

Let's calculate the ranges of projectiles A and B:

For projectile A, $v = 40 \, \text{m/s}$ and $\theta = 30^\circ$, so:

$R_A = \frac{(40)^2 \sin(2 \times 30)}{10} = 4 \times 40 = 160 \, \text{m}$.

For projectile B, $v = 60 \, \text{m/s}$ and $\theta = 60^\circ$, so:

$R_B = \frac{(60)^2 \sin(2 \times 60)}{10} = 6 \times 60 = 360 \, \text{m}$.

Therefore, the ratio of their ranges is $R_A : R_B = 160 : 360 = 4 : 9$.

$x = 3t \Rightarrow {a_x} = 0$

$y = 5{t^3} \Rightarrow {a_y} = 30t$

$\overrightarrow a (t = 1) = 30\widehat y$

Here's how to determine the ratio of the ranges for the two projectiles:

Understanding the Concepts

• Projectile Motion: Projectile motion is the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. The object is called a projectile, and its path is called its trajectory.


• Range: The range of a projectile is the horizontal distance it covers from its launch point to the point where it returns to the same vertical height.

Key Formula

The formula for the range (R) of a projectile is:

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

where:

• R = Range


• u = Initial velocity (same for both projectiles)


• θ = Angle of projection


• g = Acceleration due to gravity (constant)

Calculations

1. Projectile 1 (45° angle):

Let the range of the projectile launched at 45° be R₁.

$R_1 = \frac{u^2 \sin (2 \times 45^{\circ})}{g} = \frac{u^2 \sin 90^{\circ}}{g} = \frac{u^2}{g} $

1. Projectile 2 (30° angle):

Let the range of the projectile launched at 30° be R₂.

$R_2 = \frac{u^2 \sin (2 \times 30^{\circ})}{g} = \frac{u^2 \sin 60^{\circ}}{g} = \frac{u^2 \sqrt{3}}{2g}$

1. Ratio of Ranges (R₁ : R₂):

Divide the range of projectile 1 by the range of projectile 2:

$\frac{R_1}{R_2} = \frac{\frac{u^2}{g}}{\frac{u^2 \sqrt{3}}{2g}} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}$

To simplify the ratio, multiply both numerator and denominator by √3:

$\frac{R_1}{R_2} = \frac{2\sqrt{3} \times \sqrt{3}}{3 \times \sqrt{3}} = \frac{6}{3\sqrt{3}} = \frac{2}{\sqrt{3}}$

Therefore, the ratio of the ranges of the two projectiles is 2 : √3, which corresponds to Option C.

To solve this problem, we need to understand the relationship between the angles of projection, the initial velocities, and the time taken to reach maximum height for each projectile.

The formula to calculate the time to reach the maximum height is given by:

$ t = \frac{u \sin \theta}{g} $

where:

• $t$ is the time to reach maximum height,
• $u$ is the initial velocity,
• $\theta$ is the angle of projection, and
• $g$ is the acceleration due to gravity.

Given that the projectiles reach the maximum height in the same time, we can set up the following equation:

$ \frac{u_1 \sin 30^\circ}{g} = \frac{u_2 \sin 45^\circ}{g} $

Since $\sin 30^\circ = \frac{1}{2}$ and $\sin 45^\circ = \frac{\sqrt{2}}{2}$, the equation simplifies to:

$ \frac{u_1 \cdot \frac{1}{2}}{g} = \frac{u_2 \cdot \frac{\sqrt{2}}{2}}{g} $

Canceling out the common terms (i.e., $g$ and $\frac{1}{2}$), we get:

$ u_1 = u_2 \sqrt{2} $

Hence, the ratio of their initial velocities is:

$ \frac{u_1}{u_2} = \sqrt{2} $

Therefore, the correct answer is Option C: $\sqrt{2}:1$.

To solve this problem, we will use the equations for the range and maximum height of a projectile. The range $R$ and maximum height $H$ of a projectile launched with speed $u$ at an angle $\theta$ can be expressed as follows:

Range:

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

Maximum height:

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

We are given that the range and maximum height are equal. So, we set these equations equal to each other:

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

We can cancel out the common terms $u^2$ and $g$ on both sides:

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

Using the double angle identity for sine, $\sin(2\theta) = 2 \sin(\theta) \cos(\theta)$, we substitute it into the equation:

$2 \sin(\theta) \cos(\theta) = \frac{1}{2} \sin^2(\theta)$

We can simplify this by dividing both sides by $\sin(\theta)$ (assuming $\theta \neq 0$):

$2 \cos(\theta) = \frac{1}{2} \sin(\theta)$

Rearranging to get all terms on one side gives us:

$4 \cos(\theta) = \sin(\theta)$

Dividing both sides by $\cos(\theta)$, we get:

$4 = \tan(\theta)$

So, we find that:

$\tan(\theta) = 4$

Thus, the correct answer is:

Option D: 4

At t = 20 s,

velocity of truck,

v = 0 + 5 $\times$ 20 = 100 m/s

At 20 sec a ball is dropped from the truck, so velocity of ball will be same as truck.

Velocity of truck at x-direction = 100 m/s and in y-direction = 0.

$\therefore$ Velocity of ball v_x = 100 m/s, v_y = 0

Now ball will show projectile motion where vertically downward acceleration g = 10 m/s act on the ball.

As horizontally no acceleration acting on the ball so horizontal velocity 100 m/s will remain unchanged.

Velocity of the ball when it reach the ground along y-direction after 1 sec.

${v_y} = 0 - 10 \times 1$

$\Rightarrow$ ${v_y} = - 10$ m/s

$\therefore$ Velocity of ball $(\overrightarrow v ) = 100\widehat i - 10\widehat j$

We know,

Maximum height of a projectile $(H) = {{{u^2}{{\sin }^2}\theta } \over {2g}}$

Given, Ratio of initial velocity of two projectile

${{{u_1}} \over {{u_2}}} = {{\sqrt 3 } \over {\sqrt 2 }}$

Both projectile reach the same maximum height.

$\therefore$ ${H_1} = {H_2}$

$ \Rightarrow {{u_1^2{{\sin }^2}{\theta _1}} \over {2g}} = {{u_2^2{{\sin }^2}{\theta _2}} \over {2g}}$

$ \Rightarrow u_1^2{\sin ^2}{\theta _1} = u_2^2{\sin ^2}{\theta _2}$

$ \Rightarrow {\left( {{{{u_1}} \over {{u_2}}}} \right)^2} = {\left( {{{\sin {\theta _2}} \over {\sin {\theta _1}}}} \right)^2}$

$ \Rightarrow {\left( {{{\sqrt 3 } \over {\sqrt 2 }}} \right)^2} = {\left( {{{\sin 60^\circ } \over {\sin {\theta _1}}}} \right)^2}$

$ \Rightarrow {3 \over 2} = {3 \over {4{{\sin }^2}{\theta _1}}}$

$ \Rightarrow \sin _{{\theta _1}}^2 = {1 \over 2}$

$ \Rightarrow \sin {\theta _1} = {1 \over {\sqrt 2 }} = \sin 45^\circ $

$ \Rightarrow {\theta _1} = 45^\circ $

To determine how high a person can throw a ball given the maximum range, we need to use the principles of projectile motion in physics. The maximum range of a projectile is given by the formula:

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

where:

• $ R $ is the range
• $ v_0 $ is the initial velocity
• $ \theta $ is the angle of projection
• $ g $ is the acceleration due to gravity (approximately $ 9.8 \, \text{m/s}^2 $)

The maximum range is achieved when $ \theta = 45^\circ $, thus $ \sin(2\theta) = \sin(90^\circ) = 1 $.

Given that the maximum range $ R = 100 \, \text{m} $, we can rewrite the range formula as:

$ 100 = \frac{{v_0^2 \cdot 1}}{9.8} $

Solving for $ v_0^2 $:

$ v_0^2 = 100 \times 9.8 = 980 $

Next, the maximum height $ H $ reached by the ball can be calculated using the vertical component of the velocity. The formula for the maximum height is:

$ H = \frac{{v_0^2 \sin^2(\theta)}}{2g} $

Here, for a vertical throw, $ \theta = 90^\circ $, and thus $ \sin(90^\circ) = 1 $.

Substituting the values, we get:

$ H = \frac{{v_0^2 \cdot 1}}{2 \cdot 9.8} $

Using $ v_0^2 = 980 $, we have:

$ H = \frac{980}{2 \times 9.8} = \frac{980}{19.6} = 50 \, \text{m} $

Therefore, the correct answer is:

Option B: 50 m