HELP WITH PHYSICS QUESTION PLEASE TY

Question

The mass of the Earth is 5.98 × 1024 kg .

A 4.72 kg bowling ball initially at rest is

dropped from a height of 4.62 m.

The acceleration of gravity is 9.8 m/s2 . 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.

Answer 1

From the third law, the Force of the bowling ball on the earth is the same as the earth on the bowling ball (opposite direction)

The acceleration of the earth, a = F/M_e =( m_BBg)/M_e

The speed of the earth will be related using the constant acceleration equation v = v₀ + at, but what is t? You can see that the acceleration of the earth is very small so, we can say that the time that the earth accelerates before hitting is essentially the time for the bowling ball to fall: d = 1/2 gt² or t = sqrt(2d/g)