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 ?

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

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

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

To find the mechanical energy of the projectile initially, you can use the formula:

Ep = mgh

where:
m = mass of the projectile (2.50 kg)
g = acceleration due to gravity (9.8 m/s^2)
h = height above the ground (15.0 m)

Substituting the given values into the formula:

Ep = 2.5 * 9.8 * 15.0
Ep ≈ 367.5 Joules

Therefore, the projectile has approximately 367.5 Joules of mechanical energy initially.

Now, let's move on to the second question:

To find the work done on the projectile by gravity as it falls to the ground, you can use the formula:

W = mgh

where:
m = mass of the projectile (2.50 kg)
g = acceleration due to gravity (9.8 m/s^2)
h = height from which the projectile falls (15.0 m)

Substituting the given values:

W = 2.5 * 9.8 * 15.0
W = 367.5 Joules

Therefore, the work done on the projectile by gravity as it falls to the ground is 367.5 Joules.

Moving on to the third question:

To find the speed with which the projectile hits the ground, you can use the equations of motion.

The vertical displacement can be found using the equation:

h = (1/2)gt^2

where:
g = acceleration due to gravity (9.8 m/s^2)
t = time of flight

Rearranging the equation to solve for t:

t = sqrt(2h/g)

Substituting the given values:

t = sqrt(2 * 15.0 / 9.8)
t ≈ sqrt(3.06)
t ≈ 1.75 seconds

The horizontal distance traveled can be found using the equation:

d = v*t

where:
v = initial horizontal velocity (22 m/s)
t = time of flight (1.75 s)

Substituting the given values:

d = 22 * 1.75
d ≈ 38.5 meters

Therefore, the speed with which the projectile hits the ground is approximately 38.5 m/s.