Motion in a Plane

The first step is to determine the horizontal velocity of the ball as seen by an observer on the bank.

The velocity of the boat in still water is $27 \, \text{km/h}$. Since the boat is moving downstream and the river flows at $9 \, \text{km/h}$, the actual speed of the boat relative to the bank becomes:

$ v_{\text{boat}} = 27 + 9 = 36 \, \text{km/h} $

Convert this speed from km/h to m/s by using the conversion factor $1 \, \text{km/h} = \frac{5}{18} \, \text{m/s}$:

$ v_{\text{boat}} = 36 \times \frac{5}{18} = 10 \, \text{m/s} $

This is also the horizontal velocity of the ball as seen by the observer on the bank.

Next, calculate the time the ball spends in the air. The initial vertical velocity of the ball is $10 \, \text{m/s}$. The time to reach the maximum height $t_{\text{up}}$ is given by:

$ t_{\text{up}} = \frac{v_{\text{initial}}}{g} = \frac{10}{10} = 1 \, \text{s} $

The total time of flight $t_{\text{total}}$ is twice the time to reach the maximum height:

$ t_{\text{total}} = 2 \times 1 = 2 \, \text{s} $

The horizontal range $R$ of the ball is the horizontal velocity multiplied by the total time of flight:

$ R = v_{\text{horizontal}} \times t_{\text{total}} = 10 \times 2 = 20 \, \text{m} $

Finally, convert this range into centimeters:

$ R = 20 \times 100 = 2000 \, \text{cm} $

Thus, the range of the ball as observed by an observer at rest on the bank is $2000 \, \text{cm}$.

Time to reach max. height,

${v_y} = {u_y} - gt$

$O = 30 - 10t \Rightarrow t = 3s$

We know, ${S_n} = u + {a \over 2}(2n - 1)$

So, ${h_0} = {S_1} = 30 - {{10} \over 2}\left( {2 \times 1 - 1} \right) = 30 - 5 = 25$ m

and ${h_1} = S{w_3} = 30 - {{10} \over 2}\left( {2 \times 3 - 1} \right) = 30 - 25 = 5$

Hence, ${{{h_0}} \over {{h_1}}} = {{25} \over 5} = 5$

To solve this problem, we can use the equations of motion under uniform acceleration, separately considering the horizontal and vertical motions because the two are independent of each other.

First, for the body of mass $M$ thrown horizontally with velocity $v$ from a height $H$, let's analyze its motion:

Horizontal Motion:

The horizontal distance (range) $x$ covered by the object is given by $x = v \cdot t$, where $t$ is the time taken to hit the ground.

Vertical Motion:

The time $t$ it takes for the object to hit the ground can be found using the equation of motion under gravity, $H = \frac{1}{2} g t^2$, where $g$ is the acceleration due to gravity.

For the first body:

Given $x = 100$ m and using the equation for the vertical motion to find $t$, we have:

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

Solving for $t$, we get:

$t = \sqrt{\frac{2H}{g}}$

The horizontal motion gives:

$x = v \cdot t \Rightarrow 100 = v \cdot \sqrt{\frac{2H}{g}}$

Now, considering the second body of mass $2M$ thrown at velocity $\frac{v}{2}$ from height $4H$:

For the second body:

The time $t'$ it takes for the second body to hit the ground from height $4H$ can be found by:

$4H = \frac{1}{2} g t'^2$

Solving for $t'$, we get:

$t' = \sqrt{\frac{2 \cdot 4H}{g}} = 2 \sqrt{\frac{2H}{g}}$

This is twice the time $t$ found for the first body.

The horizontal distance $x'$ covered by the second body is:

$x' = \left(\frac{v}{2}\right) \cdot t'$

Now, substituting the value of $t'$ found above, we get:

$x' = \left(\frac{v}{2}\right) \cdot 2\sqrt{\frac{2H}{g}} = v \cdot \sqrt{\frac{2H}{g}}$

But we previously found that $v \cdot \sqrt{\frac{2H}{g}} = 100$, so:

$x' = 100 \text{ m}$

Therefore, a body of mass $2M$ thrown horizontally with velocity $v/2$ from the top of a tower of height $4H$ will touch the ground at a distance of 100 meters from the foot of the tower.

To solve this problem, we first need to understand the formula that relates the maximum height $H$ reached by a projectile to its initial velocity $v_0$ and the acceleration due to gravity $g$:

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

where:

• $H$ is the maximum height,


• $v_0$ is the initial velocity of the projectile,


• $\theta$ is the angle of projection with the horizontal, and


• $g$ is the acceleration due to gravity, which is approximately $9.8 \, \mathrm{m/s^2}$.

Given that the maximum height attained by the projectile is $64 \, \mathrm{m}$, we can write:

$64 = \frac{v_0^2 \sin^2(\theta)}{2 \cdot 9.8}$

Now, we are asked to find the new maximum height if the initial velocity is halved. Let's denote the new initial velocity as $v'_0 = \frac{v_0}{2}$. Using the formula for maximum height again, we get:

$H' = \frac{{v'_0}^2 \sin^2(\theta)}{2g}$

Substituting $v'_0 = \frac{v_0}{2}$ into this equation:

$H' = \frac{(\frac{v_0}{2})^2 \sin^2(\theta)}{2g} = \frac{v_0^2 \sin^2(\theta)}{2 \cdot 4g} = \frac{1}{4} \cdot \frac{v_0^2 \sin^2(\theta)}{2g}$

Since we know the original height:

$64 = \frac{v_0^2 \sin^2(\theta)}{2 \cdot 9.8}$

Substituting this value into our equation for $H'$, we find:

$H' = \frac{1}{4} \cdot 64 = 16 \, \mathrm{m}$

Therefore, if the initial velocity of the projectile is halved, the new maximum height reached by the projectile would be $16 \, \mathrm{m}$.

The ball needs to just cross 4 steps to just hit $5^{\text {th }}$ step

Therefore, horizontal range $(R)=0.4 \mathrm{~m}$

$\mathrm{R}=\text { u.t }$

Similarly, in vertical direction

$\begin{aligned} & \mathrm{h}=\frac{1}{2} \mathrm{gt}^2 \\ & 0.4=\frac{1}{2} \mathrm{gt}^2 \\ & 0.4=\frac{1}{2} \mathrm{~g}\left(\frac{0.4}{\mathrm{u}}\right)^2 \\ & \mathrm{u}^2=2 \\ & \mathrm{u}=\sqrt{2} \mathrm{~m} / \mathrm{s} \end{aligned}$

Therefore, $\mathrm{x}=2$

To solve for the value of $\alpha$, which represents the square of the speed of the particle at the time its $x$-coordinate is $84 \mathrm{m}$, we need to first determine the time at which the particle reaches this $x$-coordinate, and then use this time to calculate its final velocity in both the $x$ and $y$ directions.

The motion of the particle in the $x$-direction can be described by the kinematic equation for uniformly accelerated motion:

$ x = x_0 + v_{0x}t + \frac{1}{2}a_xt^2 $

Given:

$ x_0 = 0 \mathrm{~m} $

$ v_{0x} = 5 \mathrm{~m/s} $

$ a_x = 3 \mathrm{~m/s}^2 $

$ x = 84 \mathrm{~m} $ (at which we need to find the speed)

Substituting these values into the kinematic equation:

$ 84 \mathrm{~m} = 0 \mathrm{~m} + (5 \mathrm{~m/s})t + \frac{1}{2}(3 \mathrm{~m/s}^2)t^2 $

Simplifying this equation, we get:

$ 0 = \frac{3}{2}t^2 + 5t - 84 $

Using the quadratic formula to solve for $t$:

$ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $

where $a = \frac{3}{2}, b = 5, $ and $c = -84$.

$ t = \frac{-5 \pm \sqrt{(5)^2 - 4(\frac{3}{2})(-84)}}{2(\frac{3}{2})} $

$ t = \frac{-5 \pm \sqrt{25 + 504}}{3} $

$ t = \frac{-5 \pm \sqrt{529}}{3} $

$ t = \frac{-5 \pm 23}{3} $

In this scenario, since we're looking for a time when the particle reaches $84 \mathrm{m}$, we only consider the positive root because time cannot be negative.

$ t = \frac{18}{3} $

$ t = 6 \mathrm{s} $

Now we have the time at which the particle's $x$-coordinate is $84 \mathrm{m}$. Next, we find final velocities in $x$ and $y$ directions at $ t = 6 \mathrm{s} $.

The final velocity in the $x$-direction can be found using the formula for velocity with constant acceleration:

$ v_x = v_{0x} + a_xt $

$ v_x = 5 \mathrm{~m/s} + (3 \mathrm{~m/s}^2)(6 \mathrm{s}) $

$ v_x = 5 \mathrm{~m/s} + 18 \mathrm{~m/s} $

$ v_x = 23 \mathrm{~m/s} $

Similarly, for the $y$-direction:

$ v_y = v_{0y} + a_yt $

Since the particle starts from the origin and is only subject to a force after $t=0$, its initial velocity in the $y$-direction is $0$.

$ v_y = 0 + (2 \mathrm{~m/s}^2)(6 \mathrm{s}) $

$ v_y = 12 \mathrm{~m/s} $

Now we can compute the speed of the particle, which is the magnitude of the velocity vector:

$ v = \sqrt{v_x^2 + v_y^2} $

$ v = \sqrt{(23 \mathrm{~m/s})^2 + (12 \mathrm{~m/s})^2} $

$ v = \sqrt{529 + 144} $

$ v = \sqrt{673} $

Therefore, the speed of the particle at the time its $x$-coordinate is $84 \mathrm{m}$ is $\sqrt{673} \mathrm{m/s}$.

So, $\alpha = 673$.

Given:

1. The angle of projection $\theta = 30^{\circ}$.
2. The projectile is at the same height at time $t_1 = 3 \mathrm{~s}$ and $t_2 = 5 \mathrm{~s}$.
3. The acceleration due to gravity $g = 10 \mathrm{~m/s^2}$.

We need to find the initial speed of projection, $u$.

We can use the following equation to find the vertical displacement, $y$, at any time $t$:

$y = u_yt - \frac{1}{2}gt^2$

Where $u_y$ is the initial vertical component of the velocity, $u_y = u \sin \theta$.

Since the projectile is at the same height at $t_1$ and $t_2$, we can write:

$u_yt_1 - \frac{1}{2}gt_1^2 = u_yt_2 - \frac{1}{2}gt_2^2$

Substitute the values of $t_1$ and $t_2$:

$u_y(3) - \frac{1}{2}(10)(3)^2 = u_y(5) - \frac{1}{2}(10)(5)^2$

Now, let's find the initial vertical component of the velocity, $u_y$:

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

Substitute $u_y$ in the equation:

$\frac{1}{2}u(3) - \frac{1}{2}(10)(3)^2 = \frac{1}{2}u(5) - \frac{1}{2}(10)(5)^2$

Now, simplify and solve for $u$:

$3u - 90 = 5u - 250$

$2u = 160$

$u = 80 \mathrm{~m/s}$

The initial speed of projection is $80 \mathrm{~m/s}$.

When two bodies are projected from the ground at the same speed of $40 \mathrm{~ms}^{-1}$ but at different angles, and they achieve the same range, we can derive the following:

Given that one projectile is launched at an angle of $60^\circ$ with respect to the horizontal, let's denote the angles of projection as $\theta_1$ and $\theta_2$. For the ranges to be equal, we know:

$ \theta_1 + \theta_2 = 90^\circ $

Since $\theta_1 = 60^\circ$, we can find $\theta_2$ as:

$ \theta_2 = 30^\circ $

Next, to find the sum of the maximum heights attained by both projectiles, we use the formula for the maximum height $H_{\max}$:

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

For the first body projected at $60^\circ$:

$ \left(H_{\max}\right)_1 = \frac{(40)^2 \sin^2 60^\circ}{2 \times 10} $

Since $\sin 60^\circ = \frac{\sqrt{3}}{2}$:

$ \left(H_{\max}\right)_1 = \frac{1600 \times \left(\frac{\sqrt{3}}{2}\right)^2}{20} = \frac{1600 \times \frac{3}{4}}{20} = \frac{1200}{20} = 60 \mathrm{~m} $

For the second body projected at $30^\circ$:

$ \left(H_{\max}\right)_2 = \frac{(40)^2 \sin^2 30^\circ}{2 \times 10} $

Since $\sin 30^\circ = \frac{1}{2}$:

$ \left(H_{\max}\right)_2 = \frac{1600 \times \left(\frac{1}{2}\right)^2}{20} = \frac{1600 \times \frac{1}{4}}{20} = \frac{400}{20} = 20 \mathrm{~m} $

Therefore, the sum of the maximum heights attained by both projectiles is:

$ \left(H_{\max}\right)_1 + \left(H_{\max}\right)_2 = 60 \mathrm{~m} + 20 \mathrm{~m} = 80 \mathrm{~m} $

$\therefore$ ${u_1} = {u_2}\sin \theta $

${{{H_1}} \over {{H_2}}} = {{{{u_1^2} \over {2g}}} \over {u_2^2{{{{\sin }^2}\theta } \over {2g}}}}$

$ = {\left( {{{{u_1}} \over {{u_2}\sin \theta }}} \right)^2} = 1$

If the initial velocity in the horizontal direction of a projectile is represented by the unit vector $\hat{i}$ and the equation of the trajectory is given by $y = 5x(1 - x)$, we need to find the $y$ component vector of the initial velocity. (Given: $g = 10 \mathrm{\ m/s^2}$)

The trajectory equation can be expanded as:

$y = 5x - 5x^2$

In the general form of a projectile's trajectory: $y = x \tan \theta - \frac{1}{2} \frac{g x^2}{v_0^2}$

Here, the equation compares as follows:

$\tan \theta = 5 = \frac{u_y}{u_x}$

Given that the initial horizontal velocity component, $u_x$, is 1 (unit vector $\hat{i}$), we can find $u_y$ from the relationship:

$u_y = 5 \times 1 = 5$

Therefore, the $y$ component vector of the initial velocity is 5$\hat{j}$.

$ \Rightarrow v\cos \alpha = u\cos 45^\circ $ ..... (i)

& $v\sin \alpha = u\sin 45^\circ - gt$ ..... (ii)

Solve for u we get

$u = 20\sqrt 2 $ m/s

$ \Rightarrow H = {{{u^2}{{\sin }^2}45^\circ } \over {20}} = 20$ m

The situation is depicted in the following figure.

where, V_MR = velocity of man = 12 km/h

and v_R = velocity of water flow in river = 6 km/h

As, v_MR should be along CD.

$ \Rightarrow {v_R} - {v_{MR}}\sin \theta = 0$

$ \Rightarrow 6 - 12\sin \theta = 0 \Rightarrow \sin \theta = {6 \over {12}}$

$ \Rightarrow \sin \theta = {1 \over 2}$

$ \Rightarrow \theta = {\sin ^{ - 1}}\left( {{1 \over 2}} \right) = {\sin ^{ - 1}}(\sin 30^\circ )$ [$\because$ $\sin 30^\circ = {1 \over 2}$]

$ \Rightarrow \theta = 30^\circ $

$\therefore$ $\alpha = 90^\circ + \theta = 90^\circ + 30^\circ = 120^\circ $

$ \Rightarrow \alpha = 120^\circ $

For a particle ‘A’, its position along the x-axis as a function of time $ t $ is given by:

$ x(t) = 10 + 8t - 3t^2 $

To find the velocity $ v_A $, we take the derivative of $ x(t) $ with respect to $ t $:

$ v_A = \frac{d}{dt}[10 + 8t - 3t^2] = 8 - 6t $

At $ t = 1 $ second, the velocity of particle A is:

$ \vec{v_A} = (8 - 6 \cdot 1)\hat{i} = 2\hat{i} $

For a particle ‘B’, its position along the y-axis as a function of time $ t $ is given by:

$ y(t) = 5 - 8t^3 $

To find the velocity $ v_B $, we take the derivative of $ y(t) $ with respect to $ t $:

$ v_B = \frac{d}{dt}[5 - 8t^3] = -24t^2 $

At $ t = 1 $ second, the velocity of particle B is:

$ \vec{v_B} = -24 \cdot 1^2 \hat{j} = -24\hat{j} $

The velocity of particle B relative to particle A ($ \vec{v_{B/A}} $) is calculated as:

$ \vec{v_{B/A}} = \vec{v_B} - \vec{v_A} $

$ \vec{v_{B/A}} = -24\hat{j} - 2\hat{i} $

To find the magnitude of $ \vec{v_{B/A}} $:

$ |\vec{v_{B/A}}| = \sqrt{(-24)^2 + (-2)^2} = \sqrt{576 + 4} = \sqrt{580} $

Therefore, $ v $ is:

$ v = 580 $