Answer :
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.
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
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
