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.
1 answer:
Answer:
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
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