You are standing on a ladder, 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 : 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,

From third equation of motion we have :

s is the height.

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

Hence, this is the required solution.

the tires are rotating at 0.434 rpmlinear speed=45 m/sdiameter=0.550 mradius=0.275 mnow v= rωω= v/rω=45/0.275ω=163.6 rad/sω=163.6 rad/s (1rev/2π rad) (60s/1m)ω=1563 rpm