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A ball is released from the top of a tower of height h metres. It takes T seconds to reach.the ground. What is the position of the ball in T/3 seconds? |
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Answer» The acceleration of the ball will be g. Initial velocity will be 0.in T sec. body travels h mts. by applying equations of motion we gets= ut +1/2gT2 h = 1/2gT2 ------[1] in T/3 sec h1 = 1/2gT2/9 -------[2] from [1] and [2] we get h1 = h/9 distance from point of release. therefore distance from ground is h-h/9 = 8h/9 |
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