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.....
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.