Motion in a Plane

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} $.

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} $

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 $R_A : R_B = 160 : 360 = 4 : 9$.

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

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

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

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

Given,

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

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

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

Using the formula for range.

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

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

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

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

at highest point $v=0$

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

( $s$ is height from horizontal level)

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

Given,

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

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

Using vector component resolution method,

The component of velocity along $X$-axis

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

Component of velocity along $Y$-axis

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

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

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

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

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

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

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

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

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

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

The distance beyond the well where the stone will hit

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

Given,

Each step distance is equal.

1st $\rightarrow 2$ step towards east.

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

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

4th $\rightarrow$ take right and walks.

Total steps $=20$.

Draw the given data,

At last turn, person travels

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

Thus, final direction is north west.

Given,

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

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

…(i)

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

For vertical component of motion,

$ u=0 $

using equation of motion,

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

Put this in Eq. (i)

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

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

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

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

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