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Question
A ball is dropped from a building of height 45 m. Simultaneously another ball is thrown up with a speed 40 m/s. Calculate the relative speed of the balls as a function of time.
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Solution
In motion under gravity, if the ball is released or dropped that means its initial velocity is zero. In this problem, as the ball is dropped, its initial velocity will be taken as zero. We will apply kinematic equations.
According to the problem, for the ball dropped from the building, u1 = 0, u2 = 40 m/s
The velocity of the ball after time t,
v1 = u1 – gt
v1 = – gt
And for another ball which is thrown upward,
u2 = 40 m/s
The velocity of the ball after time t,
v2 = u2 – gt = (40 – gt)
∴ The relative velocity of one ball w.r.t another ball is
v12 = v1 – v2 = – gt – [– 40 – gt]
v12 = v1 – v2 = – gt + 40 + gt = 40 m/s
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