Question

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 (
g. is 32 feet/second2, how far above the ground (h) was the hammer when you dropped it? use the formula:

Answer 1

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

Answer 2

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,$a=32\ ft/s^2$

From third equation of motion we have :

$v^2-u^2=2as$

$v=\sqrt{2as}$

s is the height.

$s=\dfrac{v^2}{2a}$

$s=\dfrac{(12\ ft/s)^2}{2\times 32\ ft/s^2}$

$s=2.25\ ft$

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

Hence, this is the required solution.