Question

# A stone is allowed to fall from the top of a tower 100 m high and at the same time, another stone is projected vertically upwards from the ground with a velocity of 25 msâˆ’1. The two stones will meet after

A
4 s
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B
0.4 s
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C
0.04 s
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D
40 s
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Solution

## The correct option is A 4 sOrdinary method: Using kinematics: Let the distance from top of the tower where the stones meet be x, For first stone taking downward as positive we have from the equation of motion S=ut+12gt2 Here, S=x and u=0 So, for the first stone the point where the two stones meet is at x=12gt2(1) Fo the second stone if we take the upward direction to be positive we have S=ut+12gt2 Here S=100−x and u=25 m/s For the second stone the point where the two stones meet is at 100−x=25t−12gt2(2) Adding (1) and (2), we get 25t=100 or t=4 s Alternative method: Using Relative velocity Initial velocity of stone 1 which is dropped from the top of a tower, u1=0 Initial velocity of stone 2 which is thrown upward from the ground, u2=+25 m/s (Take upward direction to be positive) Hence, urel=u21=u2−u1=25−0=25 m/s srel=100 m arel=g−g=0 Using second equation of motion, we get Srel=urelt+12arelt2 ⇒100=25×t+0 ∴t=4 s Hence, the two stones will meet after 4 seconds.

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