JinaWu
Answered

The mass of the Earth is 5.98 × 1024 kg. A 6.83 kg bowling ball initially at rest is dropped from a height of 2.12 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?

How do you solve this?

Verified Answer :

AbhiGhost

We can assume initial velocity of earth = 0 since it won't change our results. Hope this helps you :)

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onyebuchinnaji

The speed of the ball before it hits the ground is 6.45 m/s.

The given parameters:

  • Mass of the Earth, m = 5.98 x 10²⁴ kg
  • Mass of the bowling ball, m = 6.83 kg
  • Position of the ball, h = 2.12 m

The speed of the ball before it hits the ground is calculated by applying principle of conservation of energy as follows;

\(\frac{1}{2} mv^2 = mgh\\\\v^2 = 2gh\\\\v = \sqrt{2gh} \\\\v= \sqrt{2 \times 9.8 \times 2.12} \\\\v = 6.45 \ m/s\)

Thus, the speed of the ball before it hits the ground is 6.45 m/s.

Learn more about conservation of energy here: https://brainly.com/question/166559

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