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 acceleration 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.

v = 7.18_m/s

Explanation:

Velocity of the earth towards the ball is = velocity of the ball moving towards earth

For object in free fall, we have

Where

v = final velocity

u = initial velocity

g = acceleration due to gravity

t = time

S = height of ball above ground

v^2 = u^2 - 2*g * (-S)

= 0 + 2*9.8*2.63 = 51.55_m^2/s^2

Velocity of the ball just before it hits the ground is

v = 7.18_m/s