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Q.

An elevator without a ceiling is ascending up with an acceleration of 5 m s-2. A boy on the elevator shoots a ball in vertically upward direction from a height of 2 m above the floor of elevator. At this instant, the elevator is moving up with a velocity of 10 m s-1 and floor of the elevator is at a height of 50 m from the ground. The initial speed of the ball is 15 m s-1 w.r.t. the elevator. Consider the duration for which the ball strikes the floor of the elevator in answering following questions:The time in which the ball strikes the floor of elevator is given by The maximum height reached by the ball as measured from the ground would beThe displacement of ball w.r.t. ground during its flight is

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a

2.13 s

b

4.26 s

c

1.0 s

d

2.0 s

e

52 m

f

31.25 m

g

83.25 m

h

63.25 m

i

32.64 m

j

2 m

k

52 m

l

30.64 m

answer is , , .

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Detailed Solution

Initial velocity of ball w.r.t. ground.v→BG=v→BE+v→EG=15+10=25ms−1↑v→BE=15ms−1↑,a→BG=10ms−2↓a→BE=a→BG−a→EG=10−(−5)=15ms−2↓s→BE=2m↓. Using s=ut+12at2w.r.t. elevator frame,−2=15t−12×15t2⇒t=2.13sThe height of the ball from ground at the time of projection is 52 m. Maximum height from point of projection ish=v→BG22aBG=(25)2×10=31.25mRequired height = 52 + 31.25 = 83.25 mDisplacement of floor of elevator in 2.13 s iss1=10t+12×5×t2=32.64mSo displacement of ball w.r.t. ground, s→BG=s→BE+s→EG=−2+32.64=30.64m
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