QQuestionPhysics
QuestionPhysics
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/second², how far above the ground (h) was the hammer when you dropped it? Use the formula:
V=√2gh
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Answer
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Step 1:: To solve this problem, we need to use the given formula, which is the velocity of an object when it hits the ground after free-falling from a certain height.
where $V$ is the velocity of the object, $g$ is the acceleration due to gravity, and $h$ is the height of the object above the ground.
The formula is:
Step 2:: In this case, we are given the velocity of the hammer ($V = 1$ feet/second) and the acceleration due to gravity ($g = 1$ feet/second²).
We need to find the height ($h$) from which the hammer was dropped.
Therefore, we need to rearrange the formula to solve for $h$.
Step 3:: To solve for $h$, first, square both sides of the equation:
V^2 = 2gh
Step 4:: Next, divide both sides by $2g$:
h = \frac{V^2}{2g}
Step 5:: Now, substitute the given values into the equation:
h = \frac{8^2}{2 \times 32}
Step 6:: Calculate the value:
h = \frac{64}{64} = 1
Step 7:: Therefore, the hammer was dropped from a height of 1 foot above the ground.
Final Answer
The hammer was dropped from a height of 1 foot above the ground.
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