Question

A body falls freely undergravity from a height of 40m into the top of a platform 0.8m above the ground. It velocity on reaching the platform is_______. Take g=9.8m/s

Answer 1

The law of conservation of energy gives

$$mgh_i=\frac{mv_f^2}{2}+mgh_f$$

Thus

$$v_f=\sqrt{2g(h_i-h_f)}$$

$$=\sqrt{2\times 9.8\:\rm m/s^2\times (40\: m-0.8\: m)}=27.7\:\rm m/s$$