Question

A 0.75 kilogram apple is thrown upward from the ground. The apple reaches a height of 5.5m, what is the beginning gravitational potential energy of the Apple

Answer 1

Answer:

U(0.0m) = 0J

U(5.5m) = 40.42 J

Explanation:

The gravitational potential energy is given by the following formula:

$U=mgh$

h: height

m: mass of the apple = 0.75kg

g: gravitational acceleration = 9.8m/s^2

When the apple is at 5.5m from the ground the gravitational potential energy is:

$U=(0.75kg)(9.8m/s^2)(5.5m)=40.42\ J$

when the apple is on the ground you have:

$U=mg(0m)=0\ J$