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An open lift is moving upward with veloc...

An open lift is moving upward with velocity `10 m//s.` It has an upward acceleration of `2 m//s^2.` A ball is projected upwards with velocity `20 m//s` relative to ground. Find
(a) time when ball again meets the lift
(b) displacement of lift and ball at that instant.
(c) distance travelled by the ball upto that instant.
Take `g=10 m//s^2`

Text Solution

Verified by Experts

The correct Answer is:
A
(a) At the time when ball again meets the lift.
`s_L=s_B`
`10t+1/2 xx2xxt^2=20t-1/2 xx10t^2`
Solving this equation, we get
`t=0` and `t=5/3 s`
Ball will again meet the lift after 5/3 s.
(b) At this instant `s_L=s_B = 10xx5/3 + 1/2 xx 2xx (5/3)^2 = 175/9 m=19.4 m`
(c) For the ball u is antiparallel to a. Therefore, we will first find t_0, the time when its
velocity becomes zero.
`t_0 = |u/a| = 20/10 = 2 s`
At `t(=5/3 s)lt t_0,` distance and displacement are equal
or `d=19.4 m`
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