Question

A projectile with a mass of 2.50 kg is shot horizontally from a height of 15.0 m above a flat surface. The projectile's initial speed is 22 m/s. 1.How much mechanical energy does the projectile have initially ?
2.How much work is done on the projectile by gravity as it falls to the ground?
3.With what speed does the projectile hit the ground?
1. Ep= mgh
Ep= 2.5(9.8)(15.0)
Ep = 367.5
I have no idea how to do this.....

Answer 1

1.
If you call the potential zero at ground level then at the start the projectile has:
potential + kinetic =
m g h + (1/2) m v^2
= 2.5 * 9.81 * 15 + (1/2)(2.5)(484)
= 368 Joules potential + 605 J kinetic
= 973 Joules total

2.
we did that above, it is the loss in potential energy m g h = 368 Joules

3. All kinetic when it hits the ground
(1/2) m v^2 = 973 Joules
v = 27.9 m/s

Answer 2

In problem 1, where did you get 484? Also, how did you get the velocity?

Answer 3

v = 22
22^2 = 484

the velocity is given as 22 m/s at the start

I got the final velocity from the kinetic energy , the initial ke plus the amount gained in the fall