College

You are helping with some repairs at home. You drop a hammer, and it hits the floor at a speed of 12 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. 18.0 feet
B. 1.0 foot
C. 2.25 feet
D. 8.5 feet

Answer :

Answer:

We are given the following information:

The final velocity

=

12

v=12 feet per second.

The acceleration due to gravity

=

32

g=32 feet per second squared.

We need to find the initial height

ℎ

h.

We are provided with the formula for velocity:

=

2

ℎ

v=

2gh

​

Step 1: Square both sides to remove the square root:

2

=

2

ℎ

v

2

=2gh

Step 2: Plug in the given values for

v and

g:

1

2

2

=

2

×

32

×

ℎ

12

2

=2×32×h

144

=

64

ℎ

144=64h

Step 3: Solve for

ℎ

h:

ℎ

=

144

64

h=

64

144

​

ℎ

=

2.25

 

feet

h=2.25feet

Thus, the hammer was dropped from a height of 2.25 feet.

The correct answer is C. 2.25 feet.

Explanation:

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