High School

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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 12 feet per second. If the acceleration due to gravity ([tex]g[/tex]) is 32 feet/second\(^2\), how far above the ground ([tex]h[/tex]) was the hammer when you dropped it? Use the formula:

Answer :

This is the concept of speed, distance and time, we are required to calculate the distance that a ladder which moved at a speed of 12 ft/sec fell from given that acceleration due to gravity is 32 ft/sec^2.
acceleration,a=(v-u)/t
since the hammer began at 0 speed, then
time taken will be:
a/s
=12/32
=3/8 seconds
but distance=speed*time
thus;
distance=3/8*12
=4.5 feet


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Rewritten by : Barada

Answer : Height, s = 2.25 ft.

Explanation :

It is given that,

Initial velocity of the hammer, u = 0

Final velocity of the hammer, v = 12 ft/s

Acceleration of the hammer,[tex]a=32\ ft/s^2[/tex]

From third equation of motion we have :

[tex]v^2-u^2=2as[/tex]

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

s is the height.

[tex]s=\dfrac{v^2}{2a}[/tex]

[tex]s=\dfrac{(12\ ft/s)^2}{2\times 32\ ft/s^2}[/tex]

[tex]s=2.25\ ft[/tex]

The hammer is placed 2.25 ft above the ground when it is dropped.

Hence, this is the required solution.