High School

Dear beloved readers, welcome to our website! We hope your visit here brings you valuable insights and meaningful inspiration. Thank you for taking the time to stop by and explore the content we've prepared for you.
------------------------------------------------ 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 [tex]\( g \)[/tex] is 32 feet/second[tex]^2[/tex], how far above the ground ([tex]\( h \)[/tex]) was the hammer when you dropped it? Use the formula:

[tex] v = \sqrt{2gh} [/tex]

A. 16.0 feet
B. 1.0 foot
C. 0.5 feet
D. 0.25 feet

Answer :

To find out how high above the ground the hammer was when it was dropped, you can use the formula for the velocity of an object in free fall:

[tex]\[ v = \sqrt{2gh} \][/tex]

Here, [tex]\( v \)[/tex] represents the velocity of the object when it hits the ground, [tex]\( g \)[/tex] is the acceleration due to gravity, and [tex]\( h \)[/tex] is the height from which the object was dropped.

Given values:
- [tex]\( v = 4 \)[/tex] feet per second (the speed at which the hammer hits the floor),
- [tex]\( g = 32 \)[/tex] feet per second squared (the acceleration due to gravity).

We need to rearrange the formula to solve for [tex]\( h \)[/tex]:

1. Start with the original formula:
[tex]\[ v = \sqrt{2gh} \][/tex]

2. Square both sides to eliminate the square root:
[tex]\[ v^2 = 2gh \][/tex]

3. Solve for [tex]\( h \)[/tex] by dividing both sides by [tex]\( 2g \)[/tex]:
[tex]\[ h = \frac{v^2}{2g} \][/tex]

4. Substitute the known values into the equation:
[tex]\[ h = \frac{4^2}{2 \times 32} \][/tex]

5. Calculate the result:
[tex]\[ h = \frac{16}{64} = 0.25 \][/tex]

Therefore, the hammer was dropped from a height of 0.25 feet above the ground. So the correct answer is:

D. 0.25 feet