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 speed of 10 $m{s}^{-1}$, at an angle of 60${}^{\circ }$ to the horizontal. The boy has to move forward by 1.15 m inside the train to catch the ball back at the initial height. The acceleration of the train is

• A. 3$m{s}^{-2}$
• B. 5 $m{s}^{-2}$
• C. 8 $m{s}^{-2}$
• D. 6$m{s}^{-2}$

Answer 1

The correct option is B 5 $m{s}^{-2}$

Time of flight = $\frac{2usin\theta }{g}$

=$\frac{2\times 10\times sin60}{10}$

=$\surd 3sec$

For the motion along horizontal

$S=ut+\frac{1}{2}$ $a{t}^2$

$1.15=10cos60\times t-\frac{1}{2}a{t}^2$

solving this

$a=5m/s^2$

hence (B) is the correct answer