Physics, asked by kaliyar, 1 year ago

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?

Answers

Answered by gadakhsanket
19

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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