An elevator is moving with an upward acceleration a. A coin is dropped from rest from the roof of the elevator. After what time the coin will strike the base of the elevator?

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Solution

Analyze the motion of coin with respect to the observer standing in the elevator. As the coin releases from rest inside the elevator, its velocity with respect to ground is equal to the velocity of the elevator.

Initial relative velocity of coin w.r.t. observer in the elevator,
$[_{c o i n}]_{e l e v a t o r}$ = $[_{c o i n}]_{g r o u n d}$ - $[_{e l e v a t o r}]_{g r o u n d}$ = u - u = 0

and acceleration of coin with respect to the observer in the elevator,
$[_{c o i n}]_{e l e v a t o r}$ = $[_{c o i n}]_{g r o u n d}$ - $[_{e l a v a t o r}]_{g r o u n d}$ = -g - (+a) = -(g-a)

Now using second equation for relative motion
$[_{c o i n}]_{e l e v a t o r}$ = $[_{c o i n}]_{e l e v a t o r}$ $[_{c o i n}]_{e l e v a t o r}$ t + (1/2) $[_{c o i n}]_{e l e v a t o r}$ $t^{2}$
-h = 0 - (1/2)(g + a) $t^{2}$

This yields t = $\sqrt{\frac{2 h}{g + a}}$

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