Answer :
Final answer:
Using the formula V = √2gh and rearranging for h = V² / (2g) with V=4 feet/second and g=32 feet/second², we calculate the hammer was dropped from a height of C) 0.25 feet above the ground.
Explanation:
You are standing on a ladder, helping with some repairs at home. You drop a hammer, and it hits the floor at a speed of 4 feet per second. If the acceleration due to gravity (g) is 32 feet/second², how far above the ground (h) was the hammer when you dropped it? To find out, use the formula: V = √2gh, where V is the final velocity, g is the acceleration due to gravity, and h is the height from which the object was dropped.
Rearranging this equation to solve for h gives: h = V² / (2g). Substituting the given values (V = 4 feet/second and g = 32 feet/second²) into this formula, we get h = (4²) / (2*32) = 16 / 64 = 0.25 feet. Therefore, the hammer was dropped from a height of 0.25 feet above the ground.
