Question

A train is moving along a straight line with a constant acceleration a. A boy standing in the train throws a ball forward with a speed of $10m{{s}^{-1}}$, with respect to the train at an angle ${{60}^{o}}$ to the horizontal. The boy has to move forward by $1.15m$ inside the train to catch the ball back at the initial height. The acceleration of the train in $m{{s}^{-2}}$ is:
(A). $5$
(B). $4$
(C). $6$
(D). $1$

Answer 1

A boy throws a ball at a projectile inside the train. So we can say that the ball is moving at a velocity with respect to the train. We can resolve the velocity and apply equations of motion along the x-axis and y-axis to calculate the missing values and use them to calculate the acceleration of the train.

Formula used:
$s=ut+\dfrac{1}{2}a{{t}^{2}}$

Complete step-by-step solution:
The ball performs parabolic motion inside the train.
Given, the speed of ball is $10m{{s}^{-1}}$, it makes ${{60}^{o}}$ with the horizontal
Applying equation of motion in the vertical direction,
$s=ut+\dfrac{1}{2}a{{t}^{2}}$ ----- (1)
Here,
$s$ is the displacement
$u$ is the initial velocity
$a$ is the acceleration
$t$ is the time taken

Substituting values for motion in vertical direction, we get,
$\begin{aligned} & 0=10\sin 60t-\dfrac{1}{2}g{{t}^{2}} \\\ & 0=10\dfrac{\sqrt{3}}{2}t-10\dfrac{1}{2}{{t}^{2}} \\\ & 0=t(5\sqrt{3}-5t) \\\ & t=0,\,t=\sqrt{3}s \\\ \end{aligned}$
The range of ball is $1.15m$

Applying eq (1) for motion along the horizontal direction we get,
$\begin{aligned} & 1.15=10\cos 60\times \sqrt{3}+\dfrac{1}{2}a{{(\sqrt{3})}^{2}} \\\ & \Rightarrow 1.15=5\sqrt{3}+1.5a \\\ & \Rightarrow 1.15=8.65+1.5a \\\ & \Rightarrow -7.5=1.5a \\\ & \therefore a\approx -5m{{s}^{-2}} \\\ \end{aligned}$
Therefore, the acceleration of the train is $5m{{s}^{-2}}$ in the opposite direction to the motion of the train.

Hence, the correct option is (A).

Additional Information:
When the acceleration acting on a body is constant, we can use equations of motion in one dimension which gives us a relationship between initial velocity, final velocity, acceleration, displacement and time taken. The equations of motion are; $s=ut+\dfrac{1}{2}a{{t}^{2}}$, $v=u+at$, ${{v}^{2}}={{u}^{2}}+2as$. We can apply these equations in any dimension individually.

Note:
The negative sign indicates that the train undergoes deceleration. The time taken by the ball to come back to the ground is the time period of the projectile. The vertical acceleration acting on a body in projectile motion is the acceleration due to gravity. The velocity of the body is resolved into its components. The horizontal component is the horizontal velocity and the vertical component is the vertical velocity.