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.

Answer

A, B, C, D

Explanation

(a) When the two meet,

`s_2=s_1+7`
or `(2t)=(18t)-(4.9t^2)+7`
Solving we get, `t=3.65 s`
`s_2=2xx3.65=7.3m`
`:. Height =5+7.3`
`=12.30 m`
(b) `v_(ball)=u-g t=18-9.8xx3.65=-17.77 m//s`
`:.` Velocity of ball with respect to elevator
=velocity of ball-velocity of elevator
`=(-17.77)-(2)`
`=-19.77 m//s`
`~~-19.8 m//s`
Negative sign indicates the downward
direction.