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Numerical based on gravitation
Asked by raghavendran7 | 10 Mar, 2010, 03:20: PM

Given, H = 100 m
Let the stone thrown up be stone 1
v1 = 25 m/s
a1 = - g = - 10 m/s2 (as it is moving upward)

Let the stone released from H be stone 2
v2 = 0
a2 = g = 10 m/s2  (taking g = 10 m/s2 approximately)

When the two stones meet then,

S1+S2 = H

i.e. (v1t + 1/2 a1 t2 ) + (v2 t + 1/2 a2 t2 ) = H

On substituing the values we get,

(25xt - (1/2)10xt2 ) + (0xt + (1/2)10xt2 ) = 100

On solving we get,

25xt = 100
i.e.             t = 4 seconds.

Thus the two stones will meet after 4 seconds.

Hope this helps.
TopperLearning.com

Answered by | 23 Apr, 2010, 05:02: PM

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