My try :

My option B is not getting matched.

Answer is given as B, C.

3 votes

Best answer

Your calculation is correct so far but it is not complete.

In the vertical direction there is a gravitational field of strength $g$ acting downwards. This is equivalent to an acceleration upwards at rate $g$. The acceleration of the ball relative to the container is $a_y=3g$ upwards.

In the horizontal direction the ball accelerates at $a_x$ *relative to the container*. This is different from the acceleration of the container itself. The ball must cover twice the distance in the same time. Since $s \propto a$ and $a_y=3g$ then we must have $a_x=6g$.

The final step which you have missed is to relate $a_x$ (the acceleration of the ball relative to the container) to the acceleration $a$ of the container. As in the vertical direction we have $a_x=3a=6g$. Therefore $a=2g$.

The time for the ball to reach the edge of the container is $\sqrt{\frac{L}{3g}}$ as you calculated. So the answers are options B, C.

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