Question

A stone is launched straight up by a slingshot. Its initial speed is 25 m/s and the stone is 1.8 m above the ground when launched. a)How high above the ground does the stone rise? b)How much time elapses before the stone hits the ground?

Answer 1

a) Find the height: this will be the initial height + height reached by the stone:

$$H_\text{max}=h_0+\frac{v^2}{2g}=33.7\text{ m}.$$

b) The time consists of time required to reach the maximum height and to fall down from this height on the ground (the time backwards, therefore, will be greater cause 1.8 m more must be covered):

$$t_\text{up}=\frac vg.\\\space\\ t_\text{down}=\sqrt{\frac{2H}{g}}.\\\space\\ t=t_\text{up}+t_\text{down}=\frac vg+\sqrt{\frac{2H_\text{max}}{g}}=5.17\text{ s}.$$