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A ball is thrown vertically upwards from the top of a tower with an initial velocity of 39.2 m/s. The ball reaches the ground after 5 s. Calculate (i) the height of the tower (ii) the velocity of ball on reaching the ground. (take g=9.8 m/s2) |
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Answer» BR>2 1 at 2 =19.6×5− 2 1 ×9.8×5 2 =−24.5 m Hence, HEIGHT of TOWER =24.5m v=u+at =19.6−9.8×5=−29.4 m/s |
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