A ball is dropped from a building of height 45 m. Simultaneously another ball is thrown up with a speed 40 m/s.
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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Both are free falling. Hence, there is no acceleration of one w.r.t. another. Therefore, relative speed remains constant (=40 m/s ).
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In motion under gravity, if the ball is released or dropped, that means its initial velocity is zero. Both balls are freely falling. Hence, there is no acceleration of one w.r.t. another. Therefore, relative speed remains constant (=40 m/s).
OR
Velocity of ball dropped after time t, vd = gt (downwards)
Velocity of ball thrown after time t, vt = 40 − gt (upwards)
∴ Relative speed of balls = vdt = vtd = vd + vt
= gt + (40 - gt)
= 40 m/s.
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