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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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Answer» Solution :For the ball dropped from the building , `U _(1) =0` For the ball thrown up,` u _(2) = 40 m//s` Velocity of the dropped ball after time t, `v _(1) = u _(1) + g t` `v _(1) = g t` (downward) `""…(1)` For the ball thrown up, `u _(2) = 40 m//s` Velocity of the ball after time t `v _(2) = u _(2) -0 g t = (40 - g t) ` upward) `""...(2)` RELATIVE velocity of one ball w.r.t. another ball `= v _(1) - v _(2)` `- g t[-[- (40-g t)] (because `From (1) and (2)) = 40 m.s |
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