An elecator is going up with an upward acceleration of `1m//s^2`. At the instant when its velocity is `2 m//s`, a stone is projected upward from its floor with a speed of `2m//s` relative to the elevator, at an elevation of `30^@`.
(a) Calculate the time taken by the stone to return to the floor.
(b) Sketch the path of the projectile as observed by an observer outside the elevator.
(c) If the elevator was moving with a downward acceleration equal to g, how would the motion be altered?

Answer

A, C, D

Explanation

(a) Acceleration of stone is g or `10m//s^2` in
downward direction. Acceleration of elevator is
`1m//s^2` upwards. Therefore, relative acceleration of
stone (with respect to elevator) is `11 m//s^2`
downwards. Initial velocity is already given
relative.
(b) `T = (2u_rsin theta_r)/(a_r) = ((2)(2)sin 30^@)/(11)`
`= 0.18 s`
(c) `v_S = V_(SE) + V_(E)`


(d) In that case relative acceleration between stone
and elevator will be zero. So, with respect to
elevator path is a straight line (uniform) with
constant velocity of `20 m//s` at `30^@`. But path,
with respect to man on ground will remain
unchanged.