The mass of the Earth is 5.98 × 1024 kg . A 11 kg bowling ball initially at rest is dropped from a height of 2.63 m. The acceler
ation of gravity is 9.8 m/s 2 . What is the speed of the Earth coming up to meet the ball just before the ball hits the ground? Answer in units of m/s.
Let's assume that ground level is the height 0 meters. The change in potential energy is going to be gravitational potential energy, which is given by PE=mgh. ΔPE=mgh-mgy =mg(h-y) =50(28-0) =1400 J