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 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

JEE Advanced · 2011 · 10 Apr 2011 · Paper 2 · Laws of Motion

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 m / s

, at an angle of

6 0^{\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, in

m / s^{2}

, is

Answer: 5

Solution

In the train's frame, the ball has initial horizontal velocity

v_{x} = 10 cos 60° = 5

m/s and vertical velocity

v_{y} = 10 sin 60° = 53

m/s. Time of flight:

0 = 53 t - \frac{1}{2} (10) t^{2} \Rightarrow t = 3

s. Horizontal displacement in train's frame:

x = v_{x} t - \frac{1}{2} a t^{2} = 53 - \frac{3 a}{2}

. Setting

x = 1.15

53 - \frac{3 a}{2} = 1.15 \Rightarrow a \approx 5

m/s².

JEE Advanced · 2011 · 10 Apr 2011 · Paper 2 · Integer answer

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