QuestionJune 20, 2025

A 500 kg projectile is released from a height of 210 m above the ground How deep will it penetrate the ground? Consider that while the projectile is in the air friction is negligible and while the projectile moves through the ground it is subject to a constant friction force equal to 25,000 N.

A 500 kg projectile is released from a height of 210 m above the ground How deep will it penetrate the ground? Consider that while the projectile is in the air friction is negligible and while the projectile moves through the ground it is subject to a constant friction force equal to 25,000 N.
A 500 kg projectile is released from a height of 210 m above the ground How deep will it penetrate the ground? Consider that while the projectile is in the air friction is negligible and while the projectile moves through the ground it is subject to a constant friction force equal to 25,000 N.

Solution
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Answer

41.16 m Explanation 1. Calculate Potential Energy at Height Use PE = mgh. PE = 500 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 210 \, \text{m} = 1,029,000 \, \text{J}. 2. Determine Work Done by Friction Use W = Fd. Set W = PE to find depth. 25,000 \, \text{N} \times d = 1,029,000 \, \text{J}. 3. Solve for Depth d = \frac{1,029,000 \, \text{J}}{25,000 \, \text{N}} = 41.16 \, \text{m}.

Explanation

1. Calculate Potential Energy at Height<br /> Use $PE = mgh$. $PE = 500 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 210 \, \text{m} = 1,029,000 \, \text{J}$.<br />2. Determine Work Done by Friction<br /> Use $W = Fd$. Set $W = PE$ to find depth. $25,000 \, \text{N} \times d = 1,029,000 \, \text{J}$.<br />3. Solve for Depth<br /> $d = \frac{1,029,000 \, \text{J}}{25,000 \, \text{N}} = 41.16 \, \text{m}$.
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