Motion in a Plane

$ \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} $

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} $.

Given that a boy throws a ball with an initial velocity $v_0$ at an angle $\alpha$ to the ground, we need to find the velocity at which the boy should run to catch the ball before it hits the ground.

Initial Components of the Ball's Velocity:

Horizontal Component: $v_0 \cos \alpha$

Vertical Component: $v_0 \sin \alpha$

Total Time of Flight of the Ball:

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

$ T = \frac{2 v_0 \sin \alpha}{g} $

where $g$ is the acceleration due to gravity.

Horizontal Distance Traveled by the Ball (Range $R$):

The range $R$ is calculated as:

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

Velocity of the Boy:

To catch the ball, the boy must cover the same horizontal distance as the ball in the same amount of time. Thus, the boy’s velocity $v_b$ is:

$ \begin{aligned} v_b &= \frac{R}{T} \\ &= \frac{\frac{v_0^2 \sin 2 \alpha}{g}}{\frac{2 v_0 \sin \alpha}{g}} \\ &= \frac{v_0^2 \cdot 2 \sin \alpha \cos \alpha}{2 v_0 \sin \alpha} \\ &= v_0 \cos \alpha \end{aligned} $

Therefore, the boy should run with a velocity of $v_0 \cos \alpha$ to catch the ball.

$ \begin{aligned}Given,\,\, O A & =7 \mathrm{~m} \\ u & =15 \mathrm{~m} / \mathrm{s} \end{aligned} $

Horizontal component of velocity,

$ u_x=u \cos 37^{\circ}=15 \times \frac{4}{5}=12 \mathrm{~m} / \mathrm{s} $

Vertical component of velocity,

$ \begin{aligned} & v_x=u \sin 37^{\circ} \\ & =15 \times \frac{3}{5}=9 \mathrm{~m} / \mathrm{s} \end{aligned} $

From figure, $\tan 37^{\circ}=\frac{A C}{O A}$

$ \begin{aligned} A C & =\tan 37^{\circ} \times O A \\ & =\frac{3}{4} \times 7=\frac{21}{4} \mathrm{~m}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, ...(i)\end{aligned} $

For covering horizontal distance $O A$, if $t$ is the time taken, then

$ \begin{array}{r} u_x t=7 \\ t=\frac{7}{12} \end{array} $

$ \begin{aligned} A D & =u_y t-\frac{1}{2} a t^2=9 \times \frac{7}{12}-\frac{1}{2} \times 10 \times\left(\frac{7}{12}\right)^2 \\ & =\frac{21}{4}-1.70 \end{aligned} $

From figure, $A C=A D+D C$

From Eq. (i), $\frac{21}{4}=\frac{21}{4}-1.70+y_0$

$ y_0=1.7 \mathrm{~m} $

Given the equation for the projectile's path:

$ y = 3x - 0.8x^2 $

where $g$ is the acceleration due to gravity, given as $ g = 10 \, \mathrm{m/s}^2 $.

The formula for the time of flight $ T $ is:

$ T = \frac{2u_y}{g} $

Here, $ u_y $ is the initial vertical velocity component, and $ g $ is the acceleration due to gravity.

Determining the Maximum Height

The vertex form of a parabola $(y = ax^2 + bx + c)$ offers insight into key values, with the vertex given by:

$ X_{\text{vertex}} = \frac{-b}{2a} $

For the given trajectory equation, where $ a = -0.8 $ and $ b = 3 $:

$ X_{\text{vertex}} = -\frac{3}{2 \times (-0.8)} = 1.875 \, \text{m} $

Calculating the maximum height $ y_{\max} $:

$ y_{\max} = 3 \times 1.875 - 0.8 \times (1.875)^2 = 5.625 - 0.8 \times 3.515 = 2.81 \, \text{m} $

Calculating the Initial Vertical Velocity

Using the equation for maximum height in projectile motion:

$ Y_{\max} = \frac{u_y^2}{2g} $

Solving for the initial vertical velocity $ u_y $:

$ u_y = \sqrt{2gY_{\max}} = \sqrt{2 \times 10 \times 2.81} = 7.5 \, \text{m/s} $

Calculating the Time of Flight

Finally, the time of flight is:

$ T = \frac{2u_y}{g} = \frac{2 \times 7.5}{10} = \frac{15}{10} = 1.5 \, \text{s} $

Given:

Mass of the boy, $ m = 50 \, \text{kg} $

Distance of the jump, $ d = 8 \, \text{m} $

Jump angle, $ \theta = 45^\circ $

We know the horizontal range formula in projectile motion is given by:

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

For $\theta = 45^\circ$, the equation simplifies to:

$ d = \frac{v^2}{g} $

Rearranging for $ v $, the initial velocity:

$ v = \sqrt{d \times g} $

where $ g $ (acceleration due to gravity) is:

$ g = 9.8\, \text{m/s}^2 $

Substitute the values:

$ v = \sqrt{8 \times 9.8} = 8.86 \, \text{m/s} $

Now, calculate the initial kinetic energy $ \text{KE} $:

$ \text{KE} = \frac{1}{2} \times m \times v^2 $

Substitute the known values:

$ \text{KE} = \frac{1}{2} \times 50 \times (8.86)^2 = 25 \times 78.49 = 1962.4 \, \text{J} $

Rounding this to the nearest whole number, the initial kinetic energy is approximately:

$ \text{KE} \approx 1960 \, \text{J} $

To find the percentage increase in the time of flight when the maximum height attained by a projectile is increased by 10%, we keep the angle of projection constant.

Initial Conditions:

The initial maximum height is given by the formula:

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

The time of flight is:

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

From these, we derive:

$ t^2 = \frac{4u^2 \sin^2 \theta}{g^2} = \left(\frac{8}{g}\right) h $

Therefore, $t^2 \propto h$.

With a 10% Increase in Height:

New height, $H$, is:

$ H = h + \frac{10}{100} h = \frac{11}{10} h $

Let the new time of flight be $T$. From our relationships, we have:

$ T^2 \propto H \Rightarrow \frac{T^2}{t^2} = \frac{H}{h} = \frac{11}{10} $

Simplifying:

$ \frac{T}{t} = \sqrt{\frac{11}{10}} \approx 1.05 $

The percentage increase in time of flight is:

$ \left(\frac{T-t}{t}\right) \times 100 = (1.05 - 1) \times 100 = 5\% $

Thus, the time of flight increases by 5%.

To solve the problem, we're given:

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

Horizontal range, $R = 50 \, \mathrm{m}$

First, we determine the initial velocity ($u$):

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

Plugging the values into the formula:

$ 50 = \frac{u^2 \times \sin 90^\circ}{10} $

since $\sin 90^\circ = 1$, this becomes:

$ u^2 = 500 \quad \Rightarrow \quad u = 10 \sqrt{5} \, \mathrm{m/s} $

Next, we calculate the horizontal component of the initial velocity:

$ u_{0x} = u \cos 45^\circ = 10 \sqrt{5} \times \frac{1}{\sqrt{2}} = 5 \sqrt{10} \, \mathrm{m/s} $

The horizontal displacement $x$ at time $t$ is given by:

$ x = u_{0x} \cdot t $

Given $x = 20 \, \mathrm{m}$:

$ 20 = 5 \sqrt{10} \cdot t $

Solving for $t$:

$ t = \frac{4}{\sqrt{10}} \, \mathrm{s} $

For the vertical component of the initial velocity:

$ u_{0y} = u \sin 45^\circ = 5 \sqrt{10} \, \mathrm{m/s} $

The vertical displacement $y$ at time $t$ is given by:

$ y = u_{0y} \cdot t - \frac{1}{2} g t^2 $

Substituting the known values:

$ y = 5 \sqrt{10} \cdot \frac{4}{\sqrt{10}} - \frac{1}{2} \times 10 \times \left(\frac{4}{\sqrt{10}}\right)^2 $

Simplify:

$ y = 20 - \frac{16}{2} $

Thus, the vertical displacement is:

$ y = 12 \, \mathrm{m} $

Given:

Mass of the body, $ m = 1.5 \, \text{kg} $

Initial velocity towards south, $ u_s = 8 \, \text{m/s} $

Force applied towards east, $ F_e = 6 \, \text{N} $

Time, $ t = 3 \, \text{s} $

Calculating Acceleration due to the Force

The acceleration in the eastward direction, $ a_e $, is given by:

$ a_e = \frac{F_e}{m} = \frac{6}{1.5} = 4 \, \text{m/s}^2 $

Since the initial velocity of the body in the eastward direction, $ u_e = 0 $.

Calculating Displacement

Eastward Displacement ($ s_e $)

Using the formula for displacement with constant acceleration:

$ s_e = u_e \cdot t + \frac{1}{2} a_e t^2 = 0 \cdot 3 + \frac{1}{2} \cdot 4 \cdot (3)^2 = 18 \, \text{m} $

Southward Displacement ($ s_s $)

Using the formula for displacement with constant velocity:

$ s_s = u_s \cdot t = 8 \cdot 3 = 24 \, \text{m} $

Resultant Displacement

The resultant displacement, $ s $, is the combination of the eastward and southward displacements, calculated using the Pythagorean theorem:

$ s = \sqrt{s_e^2 + s_s^2} = \sqrt{18^2 + 24^2} = \sqrt{324 + 576} = \sqrt{900} = 30 \, \text{m} $

The maximum height for a projectile can be calculated using the formula:

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

For the first stone with an angle of projection $\theta$, the maximum height $h_1$ is:

$ h_1 = \frac{20^2 \sin^2 \theta}{2 \times 10} $

$ h_1 = 20 \sin^2 \theta \quad \ldots \text{(i)} $

For the second stone projected at an angle $(90^{\circ} - \theta)$, the maximum height $h_2$ is:

$ h_2 = \frac{20^2 \sin^2(90^{\circ} - \theta)}{2 \times 10} $

$ h_2 = 20 \cos^2 \theta \quad \ldots \text{(ii)} $

We are given that the second stone rises 10 m higher than the first stone:

$ h_2 - h_1 = 10 $

Substituting the expressions for $h_1$ and $h_2$, we get:

$ 20 \cos^2 \theta - 20 \sin^2 \theta = 10 $

This simplifies to:

$ \cos^2 \theta - \sin^2 \theta = 0.5 $

Utilizing the trigonometric identity:

$ \cos^2 \theta - \sin^2 \theta = \cos 2\theta $

We find:

$ \cos 2\theta = 0.5 $

The angle $2\theta$ for which $\cos 2\theta = 0.5$ is $60^{\circ}$. Thus:

$ 2\theta = 60^{\circ} $

$ \theta = 30^{\circ} $

Given the following information:

Mass of the object: $m$

Initial velocity: $u$

Angle with the horizontal: $\theta$

Let's analyze the situation step-by-step.

Maximum Height of the Projectile

The maximum height $ H_{\text{max}} $ that the object can reach is given by:

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

Calculation of Power

Power is defined as the work done per unit of time. Therefore, the average power delivered by gravity as the object reaches its highest point can be calculated using:

$ \text{Power} = \frac{\left|\text{Work done}\right|}{\text{Time}} $

Work Done by Gravity:

The work done by gravity is equivalent to the change in potential energy when the object reaches the highest point:

$ W = \Delta U = -mgh_{\text{max}} $

Substituting for $ H_{\text{max}} $:

$ W = -mg \left(\frac{u^2 \sin^2 \theta}{2g} \right) = -\frac{mgu^2 \sin^2 \theta}{2g} $

Time to Reach Maximum Height:

To determine the time taken to reach the maximum height, you can use the vertical component of the initial velocity:

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

Average Power Delivered by Gravity:

Combine these into the power expression:

$ \text{Power} = \frac{\frac{mgu^2 \sin^2 \theta}{2g}}{\frac{u \sin \theta}{g}} $

Simplifying the expression:

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

Thus, the average power delivered by gravity while the object reaches its maximum height is:

$ P = \frac{m g u \sin \theta}{2} $

When a projectile is launched, it can achieve the same range $ R $ with two different angles of projection that are complementary, meaning they add up to $ 90^\circ $. The initial velocities for these two scenarios remain the same.

Consider the angles of projection to be $ \theta $ and $ 90^\circ - \theta $.

Time of Flight

For the two complementary angles:

Time of flight for the first angle ($ \theta $):

$ T_1 = \frac{2u \sin \theta}{g} $

Time of flight for the second angle ($ 90^\circ - \theta $):

$ T_2 = \frac{2u \cos \theta}{g} $

Product of Times of Flight

The product of the times of flight $ T_1 $ and $ T_2 $ is calculated as follows:

$ T_1 \cdot T_2 = \left(\frac{2u \sin \theta}{g}\right) \left(\frac{2u \cos \theta}{g}\right) = \frac{4u^2 \sin \theta \cos \theta}{g^2} $

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

$ T_1 \cdot T_2 = \frac{2u^2 \sin 2\theta}{g^2} $

Relationship with Range

Given that the range $ R $ is:

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

We can substitute for $ \sin 2\theta $ in terms of $ R $:

$ T_1 \cdot T_2 = \frac{2R}{g} $

Thus, the product of the times of flight $ T_1 \cdot T_2 $ is directly proportional to the range $ R $:

$ T_1 \cdot T_2 \propto R $

The time of flight for any projectile motion is given by the equation:

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

It is important to note that the time of flight does not depend on the mass of the objects involved.

Given that the times of flight for both balls are equal, we have:

$ T_1 = T_2 $

Therefore:

$ \frac{2u_1 \sin \theta}{g} = \frac{2u_2 \sin \theta}{g} $

This simplifies to:

$ u_1 = u_2 \tag{i} $

The maximum height reached by a projectile is defined by:

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

From the equation, we see that:

$ H \propto u^2 $

Given $ u_1 = u_2 $ from equation (i):

$ \frac{H_1}{H_2} = \frac{u_1^2}{u_2^2} = \frac{u_1^2}{u_1^2} = 1 $

Thus, the ratio of their maximum heights is:

$ H_1:H_2 = 1:1 $

In a sports event, a disc is thrown to achieve a maximum range of 80 meters. Our task is to determine the distance the disc travels in the first 3 seconds, given that the acceleration due to gravity ($g$) is $10 \, \text{m/s}^2$.

Let's start by understanding the maximum range:

$ R_{\max} = 80 \, \text{m} $

For maximum range in projectile motion, the launch angle is 45 degrees. The equation for maximum range is:

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

Substituting $\theta = 45^\circ$:

$ \frac{u^2 \sin(2 \times 45^\circ)}{g} = 80 $

Since $\sin 90^\circ = 1$, we have:

$ \frac{u^2}{g} = 80 $

Solving for $u^2$:

$ u^2 = 80 \times 10 = 800 $

Thus, the initial velocity ($u$) is:

$ u = \sqrt{800} = 20\sqrt{2} \, \text{m/s} $

Next, let's find the horizontal component of the velocity:

$ u_x = u \cos 45^\circ = 20\sqrt{2} \times \frac{1}{\sqrt{2}} = 20 \, \text{m/s} $

The vertical component of the velocity is:

$ u_y = u \sin 45^\circ = 20\sqrt{2} \times \frac{1}{\sqrt{2}} = 20 \, \text{m/s} $

Now, calculate the horizontal distance traveled in the first 3 seconds:

$ s_x = u_x \times t = 20 \, \text{m/s} \times 3 \, \text{s} = 60 \, \text{m} $

Therefore, the distance traveled by the disc in the first 3 seconds is 60 meters.

To analyze the problem, consider two objects:

First Object: Maximizing Range

To achieve maximum range, the object must be projected at a $45^\circ$ angle.

Using the formula for maximum height $H_1$ when launched at an angle $\theta_1 = 45^\circ$:

$ H_1 = \frac{u_1^2 \sin^2 \theta_1}{2g} = \frac{u_1^2 \times 1}{2g \times 2} = \frac{u_1^2}{4g} \quad \text{(i)} $

Second Object: Maximizing Height

To achieve maximum height, the object should be projected vertically at a $90^\circ$ angle.

The formula for the maximum height $H_2$ with a launch angle $\theta_2 = 90^\circ$:

$ H_2 = \frac{u_2^2 \sin^2 90^\circ}{2g} = \frac{u_2^2}{2g} \quad \text{(ii)} $

Equating Heights

Since both objects reach the same maximum height:

$ H_1 = H_2 $

From equations (i) and (ii):

$ \frac{u_1^2}{4g} = \frac{u_2^2}{2g} $

Simplifying, we find:

$ u_1^2 = 2u_2^2 $

Taking the square root:

$ u_1 = \sqrt{2}u_2 $

Thus, the ratio of their initial velocities is:

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

Therefore, the ratio of initial velocities is $\sqrt{2}:1$.

Given.

Angle of projection $=45^{\circ}$ height of wall $=$ h

We know that.

The equation of trajectory of a projectale,

$ \begin{aligned} & y=x \tan \theta-\frac{g x^2}{2 u^2 \cos ^2 \theta} \\ & y=x \tan \theta\left(1-\frac{g x}{2 u^2 \cos ^2 \theta \cdot \tan \theta}\right) \\ & y=x \tan \theta\left(1-\frac{g x}{2 u^2 \cos ^2 \theta \cdot \frac{\sin \theta}{\cos \theta}}\right) \\ & y=x \tan \theta\left(1-\frac{x}{R}\right)\left\{\because R=\frac{u^2 \sin 2 \theta}{g}\right\} \end{aligned} $

Here, $R=d_1+d_2, x=d_1$ and $y=h$

$ \begin{aligned} \text { So, } & h=d_1 \tan 45^{\circ}\left(1-\frac{d_1}{d_1+d_2}\right) \\ & h=d_1\left(\frac{d_2}{d_1+d_2}\right) \\ & h=\frac{d_1 d_2}{d_1+d_2} \end{aligned} $

Let in is body reaches upto point $A$ and after two more second, it reaches upto point $B$

Total time of ascent for a body is,

$ \begin{aligned} t_{\text {ascent }} & =2+1=3 \mathrm{~s} \\ t_{\text {ascent }} & =\frac{\text { Time of flight }}{2}=\frac{2 u \sin \theta}{2 \times g} \\ \frac{u \sin \theta}{g} & =3 \Rightarrow u \sin 45^{\circ}=30 \quad \ldots \text { (i) } \end{aligned} $

Horizontal component of velocity remains always constant,

$ u \cos 45^{\circ}=v \cos 45^{\circ} \quad \ldots \text { (ii) } $

For vertical upward motion between point $O$ and $A$,

$ \begin{aligned} & v_y=u_y-g t \\ & v \sin 45^{\circ}=u \sin 45^{\circ}-10 \times 1 \\ & \frac{v}{\sqrt{2}}=30-10 \quad \text { from Eq. (i) } \\ & \quad v=20 \sqrt{2} \mathrm{~m} / \mathrm{s} \end{aligned} $

Putting this value in Eq. (ii),

$ u \cos 45^{\circ}=20 \sqrt{2} \times \frac{1}{\sqrt{2}}=20 \quad \ldots \text { (iii) } $

From Eq. (i) and Eq. (iii),

$ \begin{aligned} & u^2 \sin ^2 45^{\circ}+u^2 \cos ^2 45^{\circ}=(30)^2+(20)^2 \\ & u^2=1300 \\ & u=10 \sqrt{13} \mathrm{~m} / \mathrm{s} \end{aligned} $

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

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

where:

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

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

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

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

$v_0^2 = R \cdot g$

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

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

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

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

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

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

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

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

The given situation is shown below

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

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

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

$\because$ Magnitude of resultant force,

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

Magnitude of acceleration of the object

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

The given situation can be described by the figure given below

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

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

Angle of projection,

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

Maximum height attained by projectile,

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Given, friction force due to air = $f_a$

According to given diagram

Let W be the weight of ball.

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

$\therefore$ Free body diagram of ball will be

Given, mass of particle $=m$

Velocity $=u$

Angle of projection $=\theta$

Let angular momentum, $L=m v r$

Maximum height $=H$

As we know that,

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

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

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

where:

T = time of flight

u = initial velocity

θ = angle of projection

g = acceleration due to gravity

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

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

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

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

$T = 5 s$

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

So the answer is Option A.