QQuestionPhysics
QuestionPhysics
"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/second^2 , how far above the ground (h) was the hammer when you dropped it? Use the formula
V= square root 2gh
A. 16.0 feet
B. 1.0 foot
C. 0.25 feet
D. 0.5 feet"
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Answer
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Step 1:I'll solve this problem step by step, following the specified LaTeX formatting guidelines:
Step 2:: Identify the Given Information
- Acceleration due to gravity (g) = $$32 \frac{\mathrm{feet}}{\mathrm{second}^{2}}
- Final velocity (V) = 4 feet/second - We need to find the initial height (h)
Step 3:: Rearrange the Equation
h = \frac{V^{2}}{2g}
To solve for h, we'll rearrange the equation:
Step 4:: Substitute Known Values
h = \frac{(4 \mathrm{~feet/second})^{2}}{2 \cdot 32 \mathrm{~feet/second}^{2}}
Step 5:: Calculate the Height
h = \frac{16 \mathrm{~feet}^{2}/\mathrm{second}^{2}}{64 \mathrm{~feet}/\mathrm{second}^{2}}
h = 0.25 \mathrm{~feet}
Final Answer
Key Insights: - The equation V = \sqrt{2gh} relates velocity, gravity, and height - Rearranging allows us to solve for height when velocity is known - Always check units to ensure correct calculation
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