A 0.245kg ball is thrown straight up from 2.20m above the ground. Its initial vertical speed is 8.80m/s. A short time later, it hits the ground. Calculate the total work done by the force of gravity during that time.

Answer 1

Fg = m*g = 0.245kg * 9.8N/kg = 2.401 N.=
Force of gravity.

D = d1 + d2
d1=(V^2-Vo^2)/2g=(0-8.8^2)/-19.6=3.951 m
Up.

d2 = d1 + 2.20 = 3.951 + 2.20 = 6.15 m,
Down.

Wg = Fg*(d1+d2)=2.401*(3.951+6.15)=24.3
Joules.