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Question

# Two balls of equal masses are thrown upwards along the same vertical direction at an interval of 2s, with the same initial velocity of 39.2m/s. The two balls will collide at a height of

A
39.2m
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B
73.5m
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C
78.4m
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D
117.6m
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Solution

## The correct option is C 73.5mGiven that,Let the mass of the two body be m1=m2=mLet two balls collide at a height s from the ground after t second when second ball is thrown upwards.∴ Time taken by first ball to reach the point of collision =(t+2)sAs per the Newton's second law of motion,s=ut+12g×t2s=39.2(t+2)+12(−9.8)(t+2)2=39.2(t+2)−4.9(t+2)2 ...(i)For second balls=39.2t+12(−9.8)t2=39.2t−4.9t2 ...(ii)From eqs. (i) and (ii)39.2(t+2)−4.9(t+2)2=(39.2)t−4.9t2On solving we get, t=3sFrom Eq. (ii),s=39.2×3−4.9×(3)2=117.6−44.1=73.5mso at the height of 73.5m these two balls will collide.

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