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  • You are helping with some repairs at home. You drop a hammer and it hits the floor at a speed of 8 feet per second. If the acceleration due to gravity (g) is 32 feet/second^2, how far above the ground (h) was the hammer when you dropped it? Use the formula: v = √2gh

Question

You are helping with some repairs at home. You drop a hammer and it hits…

You are helping with some repairs at home. You drop a hammer and it hits the floor at a speed of 8 feet per second. If the acceleration due to gravity (g) is 32 feet/second^2, how far above the ground (h) was the hammer when you dropped it? Use the formula: v = √2gh

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Answer

The correct answer is: $1\text{ ft}$

Explanation

You are given $v=\sqrt{2gh}$. Solve for $h$ by squaring both sides and isolating $h$.

Steps:

  1. Square both sides: $$v^2 = 2gh$$
  2. Solve for $h$: $$h = \frac{v^2}{2g}$$
  3. Substitute $v=8\text{ ft/s}$ and $g=32\text{ ft/s}^2$: $$h = \frac{8^2}{2\times32} = \frac{64}{64} = 1\text{ ft}$$

Therefore the hammer was dropped from a height of $1\text{ ft}$.

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