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?

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?

A body travels one half of a distance with uniform velocity $v_1$ and the other half with uniform velocity $v_2$. Find the magnitude of the average velocity.

A car is moving on the road when the rain is falling vertically downwards. Why does the front windscreen get wet whereas the hind screen remains dry?

Can an object have velocity and acceleration in perpendicular directions?

If a body is thrown vertically upward from a vehicle moving with uniform velocity, where will the body fall?