Answer to Question #100098 in Mechanics | Relativity for Feker

Question

A ball is thrown horizontally from a height of 20 meter and hits the ground with a speed that is three times its initial speed what is the initial speed?

Answer 1

The law of conservation of energy gives

$$mgh+\frac{mv_i^2}{2}=\frac{mv_f^2}{2}$$

$$mgh+\frac{mv_i^2}{2}=\frac{m(3v_i)^2}{2}$$

So

$$gh=4v_i^2$$

$$v_i=\frac{\sqrt{gh}}{2}=\frac{\sqrt{9.8\:\rm m/s^2\times 20\:\rm m}}{2}=7\:\rm m/s$$