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A ball is dropped from a building of height \( 45 \mathrm{~m} \). Simultaneously another ball is thrown up with a speed \( 40 \mathrm{~m} / \mathrm{s} \). Calculate the relative speed of the balls as a function of time.
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The correct answer is:
\( 40 \mathrm{~m} / \mathrm{s} \)
For the ball dropped from building, u1 = 0
Velocity of the dropped ball after time t
v1 = u1 + gt
v1 = gt (downward)
For the ball thrown up, u2 = 40 m/s
Velocity of the ball after time t
v2 = u2 – gt
= (40 – gt) (upward)
∴ Relative velocity of one ball w.r.t another ball
= v1 – v2
= gt – [– (40 – gt) ] = 40 m/s
Velocity of the dropped ball after time t
v1 = u1 + gt
v1 = gt (downward)
For the ball thrown up, u2 = 40 m/s
Velocity of the ball after time t
v2 = u2 – gt
= (40 – gt) (upward)
∴ Relative velocity of one ball w.r.t another ball
= v1 – v2
= gt – [– (40 – gt) ] = 40 m/s
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