Question

an elastic ball strikes a moving elevator with velocity V if the elevator has a speed V the maximum height attained by the ball after collision measured from point of collision​

Answer 1

Explanation:

is 0. So the velocity with which the ball strikes the elevator would be velocity of ball plus velocity of elevator which is 2v. Elastic collision would give an equal velocity in upward direction. Now apply third kinematic equation by putting the value of v as 0 and u as 2v to get the answer as 2v^2/g

Answer 2

Given:

Velocity of an elastic ball $=V$

Speed of an elevator $=V$

To Find: The maximum height attained by the ball after collision measured from point of collision

Solution:

As the elastic ball strikes the elevator and the velocity of both ball and elevator is $V$, therefore,

The total velocity at the point of collision is given as:

$\[V + V = 2V\]$

As we know, the elastic collision would give an equal velocity in the upward direction, therefore,

By the third equation of kinematic motion,

$\[\begin{gathered} {v^2} = {u^2} - 2gh \hfill \\ - h = \frac{{{v^2} - {u^2}}}{{2g}} \hfill \\ \end{gathered} \]$

Now, after the collision, the initial velocity is $2V$ and at maximum height, the final velocity becomes $0$. Thus,

Put $u$ as $2V$ and $v$ as $0$, we have,

$\[\begin{gathered} - h = \frac{{0 - {{(2V)}^2}}}{{2g}} \hfill \\ - h = \frac{{ - 4{V^2}}}{{2g}} \hfill \\ h = \frac{{2{V^2}}}{g} \hfill \\ \end{gathered} \]$

Hence, the maximum height attained by the ball after collision measured from point of collision is $\[\frac{{2{V^2}}}{g}\]$.

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