# A body reaches a certain height when thrown with a certain velocity. What is the height reached if the velocity is doubled?

Let $h$ be the height reached by the body when thrown upwards with a velocity $u$. At maximum height, the velocity will be zero $v=0$. $\therefore v^2=u^2-2gh$ $0=u^2-2gh$ $\therefore h=\frac{u^2}{2g}$

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

Let $h’$ be the new height reached by the body, then $h’=\frac{u’^2}{2g}=\frac{4u^2}{2g}=4h$

Thus, the new height will be four times the original.

[Equations used above are deduced from Motion Under Gravity]

#### SIMILAR QUESTIONS

A body thrown vertically upwards takes ‘t’ time to reach the highest point. What would be the time required to reach the same point during its return?

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