A boy standing at the top of a tower of 80 m height drops a stone. Assuming $g=10 \mathrm{~m} \mathr

Question

A boy standing at the top of a tower of 80 m height drops a stone. Assuming $g=10 \mathrm{~m} \mathrm{~s}^{-2}$, the velocity with which it hits the ground is

Professor PC Thomas Classes · Answer

(b) : Here, $u=0, g=10 \mathrm{~m} \mathrm{~s}^{-2}, h=80 \mathrm{~m}$ \ Let $v$ be the velocity with which the stone hits the ground. \ $	herefore \quad v^{2}=u^{2}+2 g h$ \ or $v=\sqrt{2 g h}=\sqrt{2 	imes 10 	imes 80}=\sqrt{1600}=40 \mathrm{~m} / \mathrm{s}$