Q: An open elevator is ascending with constant speed of 10 m/s. A ball is thrown vertically up by a boy on elevator when it is at a height of 40 m from the ground. The velocity of projection with respect to the elevator is 30 m/s. The maximum height attained by the ball is

(take g = 10 ms^{2})

(a) 85 m

(b) 60 m

(c) 120 m

(d) 45 m

Ans:(c)

Sol: h = 40 m ; Initial vel. w.r.t. elevator = 30 m/s

Velocity of elevator = 10 m/s

Velocity w.r.t. ground, u = 30 + 10 = 40 m/s

From v^{2} = u^{2} + 2 a S

0= 1600 + 2(-10)h’

h ‘ = (-1600)/(-20) = 80 m

Total height above the ground = 40 + 80 = 120 m