Let $h$ be the greatest height reached by the body when thrown upwards with a velocity $u$. At maximum height, the velocity will be zero $(v=0).$

\[\therefore v^2=u^2-2gh\] \[0=u^2-2gh\] \[\therefore h=\frac{u^2}{2g}\]

At half the greatest height, the distance travelled by the body is $\frac{h}{2}$. Let $v’$ be the velocity at this point. \[\therefore v’^2=u^2-2g\frac{h}{2}\] \[v’^2=u^2-2g×\frac{u^2}{4g}\] \[v’^2=u^2-\frac{u^2}{2}\] \[v’^2=\frac{u^2}{2}\] \[\therefore v’=\frac{u}{\sqrt{2}}\]

Thus, the velocity at half the greatest height will be $\frac{1}{\sqrt{2}}$ times the velocity with which it was thrown upwards.

[Equations used above are deduced from Motion Under Gravity]

**SIMILAR QUESTIONS**

**Rain drops hitting the side windows of a car in motion often leave diagonal streaks. Why?**