Question

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 1

Correct Answer - A::B::C::D
(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.