# A body when thrown upwards with a certain velocity requires certain time to reach the maximum height. If the velocity is doubled, what is the time required to reach the highest point?

Let $t$ be the time required by the body to reach the maximum height $h$ when thrown upwards with a velocity $u$. At the maximum height, the velocity will be zero $v=0.$

$v=u-gt$ $0=u-gt$ $\therefore t=\frac{u}{g}$

If the velocity is doubled, then new velocity will be $u’=2u.$

Let $t’$ be the new time to reach the highest point. $t’=\frac{u’}{g}=\frac{2u}{g}=2t$

Thus, the new time required to reach the highest point will be twice the original time.

[Equations used above are deduced from Motion Under Gravity]

#### SIMILAR QUESTIONS

A body is dropped from a certain point and it reaches a velocity ‘v’ after some time. What would be the velocity reached after twice the time?

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