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.

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west