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?

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.

Two bodies of different masses are dropped from same height. Compare their velocities on reaching the ground and the time required.

A boy wants to take his boat to a point just opposite on the other bank of the river with water flowing its usual course. How should he row and why?

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

Can an object have velocity and acceleration in perpendicular directions?

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