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 ms2)
(a) 85 m
(b) 60 m
(c) 120 m
(d) 45 m
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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 v2 = u2 + 2 a S
0= 1600 + 2(-10)h’
h ‘ = (-1600)/(-20) = 80 m
Total height above the ground = 40 + 80 = 120 m