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Question

A ball is dropped from the top of a building and simultaneously, another ball is projected vertically upward from the ground with a velocity of 25 m/s. If the initial distance between these balls is 100 m. Calculate where the two balls will meet above the ground.
(Take g = 10 m/s2 )

A
20 m
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B
40 m
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C
60 m
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D
80 m
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Solution

The correct option is A 20 m
We consider that the ball dropped from the building is b1 and another ball, projected vertically upward is b2.
Both the balls have an acceleration of g in the downward direction and they meet after ‘t’ sec.

Initial velocity of ball(b1) = 0 m/s

Initial velocity of ball(b2) = 25 m/s

Distance covered by ball(b2) in time t = x

Total height of the cliff(H) = 100 m

●Distance covered by ball(b1) in time t = 100 x

Plug in values in second equation of motion

For Ball (b1) :

100 x = 0 + 12gt2

100 x = 12gt2 ....(1)

For Ball(b2) :

x = u2t - 12gt2 …(2)

On adding (1) & (2), we get the final equation :

100 =12gt2 +u2t - 12gt2

100 = u2t

t = 4 s

Substituting t = 4 in equation 1

x = (25)(4) - 12(1042)

x = 20 m


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