JEE Advance - Physics (2011 - Paper 2 Offline - No. 7)
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 $$60^\circ $$ 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 m/s2, is
Answer
5
Explanation
u = 10 ms$$-$$1, $$\theta$$ = 60$$^\circ$$
Time of flight is
$$t = {{2u\sin \theta } \over g} = {{2 \times 10 \times \sin 60^\circ } \over {10}} = \sqrt 3 \,s$$
Let v be the velocity of the train. The horizontal velocity of ball at the instant it is thrown $$ = (v + {u_x}) = (v + u\cos \theta )$$. Therefore, the horizontal range of the ball with respect to the ground is
$$R = (v + u\cos \theta )t$$, where $$t = \sqrt 3 \,s$$
It is clear that
Distance travelled by ball in time $$t + 1.15 = R$$
i.e. $$vt + {1 \over 2}a{t^2} + 1.15 = (v + u\cos \theta )t$$
$$ \Rightarrow {1 \over 2}a{t^2} + 1.15 = (u\cos \theta )t$$
$$ \Rightarrow {1 \over 2}a \times {(\sqrt 3 )^2} + 1.15 = (10\cos 60^\circ ) \times \sqrt 3 $$
$$ \Rightarrow a = 5$$ ms$$-$$2
Comments (0)
