A ball is thrown vertically upward from the 12 m level with an initial velocity of `18 m//s.` At the same instant an open platform elevator passes the 5 m level, moving upward with a constant velocity of `2 m//s.` Determine (`g = 9.8 m//s^2` )
(a) when and where the ball will meet the elevator,
(b) the relative velocity of the ball with respect to the elevator when the ball hits the elevator.

Options

A. `10.2m 9.8m//s`
B. `12.3m 19.8m//s`
C. `12m 10.2m//s`
D. `12.5m 22m//s`

Answer

B

Explanation

a) Let the two meet at a distance `s` from ground.
Then `s-12=18t-1/2xx9.8xxt^(2)`..(i)
and `s-5=2t`…(ii) Solving these two equations, we get `t=3.65s` and `s=12.30m`
b) `v_(b)=18-(9.8)(3.65)=-17.8m//s` i.e, velocity of ball is `17.8 m//s` (downward) at the time of impact or relative velocity `=19.8 m//s` (downward)