A ball is dropped from a building of height 45 m. Simultaneously another ball is thrown up with a speed 40 m/s. The relative speed of the balls varies with time as

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$t^{2}$

$t^{0}$

$t^{1}$

1/t

answer is A.

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Detailed Solution

According to the problem, for the ball dropped from the building, u₁=0 and u₂=40m/s

Velocity of the ball after time t,

v₁ = u₁− gt

v₁= −gt

And for another ball which is thrown upward,

u₂= 40m/s

Velocity of the ball after time t,

v₂ = u₂− gt = (40 − gt)

Therefore,

Relative velocity of one ball with respect to another ball is,

v₁₂ = v₁ − v₂ = −gt − [−40 − gt]

v₁₂ = v₁ − v₂ = −gt + 40 + gt = 40m/s

Therefore, relative speed is indepent of time.

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