Question

A projectile is launched straight up at 60.0 m/s from a height of 80.0 m, at the edge of a sheer cliff. The projectile falls, just missing the cliff and hits the ground below. a. What is the maximum height of the projectile above the point of firing? b. How long does it take to hit the ground below the cliff? c. What is the projectile’s velocity upon impact with the ground?

Answer 1

a.

$$h=\frac{v^2}{2g}=\frac{60^2}{2(9.8)}=184 m$$

b. Motion angled upwards

$$t_1={\frac{v}{g}}={\frac{60}{9.8}}=6.12 s.$$

Motion angled downwards

$$t_2=\sqrt{\frac{2(H+h)}{g}}=\sqrt{\frac{2(80+184)}{9.8}}=7.34 s.$$

$$t=t_1+t_2=6.12+7.34=13.5 s$$

c.

$$V=\sqrt{v^2+2gH}=\sqrt{60^2+2(9.8)(80)}=71.9 \frac{m}{s}$$