When the ball is thrown vertically upwards with velocity $u,$ at maximum height $h,$ its velocity will be zero $v=0$. \[v^2 = u^2 – 2gh\] \[0 = u^2 – 2gh\] \[\therefore u = \sqrt{2gh}\]

The ball comes down to the earth with initial velocity zero $u’=0$ and final velocity $v’$. \[v’^2=u’^2+2gh\] \[v’^2 = 0 + 2gh \] \[\therefore v’ = \sqrt{2gh}\]

Thus, we get, \[u = v’ = \sqrt{2gh}\]

Hence, the ball comes down to the earth with the velocity with which it was thrown vertically upwards.

[Equations used above are deduced from Motion Under Gravity]

**Alternative Method**

When the ball is thrown vertically upwards with velocity $u$, it reaches the maximum height; its velocity will be zero and the ball returns to the earth. When it reaches the earth, the vertical displacement is zero, which means $h=0$. If $v$ be the velocity with which the ball reaches down to the earth, then

\[v^2=u^2-2gh\] \[v^2=u^2-2g(0)\] \[v^2=u^2\] \[\therefore v = u\]

Hence, the ball comes down to the earth with the velocity with which it was thrown vertically upwards.

[Equations used above are deduced from Motion Under Gravity]

**SIMILAR QUESTIONS**

**Can an object have an eastward velocity while experiencing a westward acceleration?**