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JEE Advanced · Physics · 4. Motion in 2D

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 \(10 \mathrm{~m} / \mathrm{s}\), at an angle of \(60^{\circ}\) to the horizontal. The boy has to move forward by \(1.15 \mathrm{~m}\) inside the train to catch the ball back at the initial height. The acceleration of the train in \(\mathrm{m} / \mathrm{s}^2\), is

  1. A 2
  2. B 3
  3. C 5
  4. D 7
Verified Solution

Answer & Solution

Correct Answer

(C) 5

Step-by-step Solution

Detailed explanation

\(\begin{aligned} t & =T=\frac{2 u \sin \theta}{g} \\ & =\frac{2 \times 10 \times \sin 60^{\circ}}{10}=\sqrt{3} \mathrm{~s}\end{aligned}\)
Displacement of train in time \(t=\frac{1}{2} a t^2\)
Displacement of boy with respect to train
\(
=1.15 \mathrm{~m}
\)
\(\therefore\) Displacement of boy with respect to ground \(=\left(1.15+\frac{1}{2} a t^2\right)\)
Displacement of ball with respect to ground \(=\left(u \cos 60^{\circ}\right) t\)
To catch the ball back at initial height,
\(
\begin{aligned}
1.15+\frac{1}{2} a t^2 & =\left(u \cos 60^{\circ}\right) t \\
\therefore 1.15+\frac{1}{2} a(\sqrt{3})^2 & =10 \times \frac{1}{2} \times \sqrt{3}
\end{aligned}
\)
Solving this equation, we get
\(
a=5 \mathrm{~ms}^{-2}
\)
\(\therefore\) Answer is 5 .
Analysis of Question
(i) Question is moderately tough.
(ii) Velocity of ball given in the question is with respect to ground.
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