Question

question 10 of 10
you are 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? use the formula:

v = √(2gh)

a. 1.0 foot
b. 0.5 feet
c. 0.25 feet
d. 16.0 feet

Explanation:

Step1: Start with the given formula

We have the formula \( v = \sqrt{2gh} \), and we need to solve for \( h \). First, square both sides of the equation to eliminate the square root.
\( v^2 = 2gh \)

Step2: Solve for \( h \)

Rearrange the formula to solve for \( h \) by dividing both sides by \( 2g \). So, \( h=\frac{v^2}{2g} \)

Step3: Substitute the given values

We know that \( v = 4 \) feet per second and \( g = 32 \) feet/second². Substitute these values into the formula for \( h \).
\( h=\frac{4^2}{2\times32} \)

Step4: Calculate the numerator and denominator

First, calculate the numerator: \( 4^2 = 16 \). Then, calculate the denominator: \( 2\times32 = 64 \).

Step5: Divide to find \( h \)

Now, divide the numerator by the denominator: \( h=\frac{16}{64}=\frac{1}{4} = 0.25 \) feet.

Answer:

C. 0.25 feet