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]


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