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A boy standing on a stationary lift (open from above) throws a ball upwards with the maximum initial speed he can, equal to 49 ms-1. How much time does the bail take to return to his hands? if the lift starts moving up with a uniform speed of 5 ms-1 and the boy again throws the ball up with the maximum speed he can, how long does the ball take to return to his hands? |
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Answer» Initial velocity n = 49 ms-1 Acceleration g = -9.8 ms2 Displacement s = 0 s = ut + 1/2 at2 0 = 49 t + 1/2 (-9.8) t2 ⇒ t = 0 or t = 10 s t = 0 represents initial time. The required ‘t’ here is t = 10 s. In the case where the lift also moves, the relative velocity of the ball with respect to the boy remains the same, i.e, 49 ms-1. The relative displacement is also 0. Hence the time required in this case is also 10 s. |
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