Question

A stone is thrown vertically upward with a speed of 15.0 m/s from the edge of a cliff 75.0 m high.

(a).How much later does it reach the bottom of the cliff?

(b).What is its speed just before hitting?

(c).What total distance did it travel?

Answer 1

a)

$$y=0=h+vt-0.5gt^2$$

$$0=75+15t-0.5(9.8)t^2$$

$$t=5.73\ s$$

b)

$$V^2=v^2+2gh$$

$$V^2=15^2+2(9.8)(75)$$

$$V=41.2\frac{m}{s}$$

c)

$$s=h+2\frac{v^2}{2g}=h+\frac{v^2}{g}$$

$$s=75+\frac{15^2}{9.8}=98.0\ m$$