An object projected up vertically with a velocity of 98 m/s reaches a point X in its path after 4 seconds. It reaches X again after another how many seconds?
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Hey Dear,
◆ Answer -
16 s after throwing
◆ Explaination -
Let t be the time required for object reach maximum height.
v = u + at
0 = 98 + (-9.8)×t
t = 98 / 9.8
t = 10 s
It's given that object reaches X at 4 s after which it travels further 10-4 = 6 s to reach max height.
Object will take same time i.e. 6 s to return from maximum height to X.
Thus, the object reaches at X again after 10+6 = 16 s from time it was thrown.
Thanks dear...
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