A person is standing at the top of a 58.0m tall hill. He stumbles and falls.As he rolls down the hill, magically friction has no effect on him! When he is still 25.0m above the base of the hill determine how fast he is moving.
If friction has no effect, why would he roll and not just slide?
I will assume in fact that there is no friction and he does not roll so translational kinetic energy is all the kinetic energy he has (no rotational)
58-
change in potential energy = m g (25 -58)
change in kinetic energy = (1/2) m v^2
so
m v^2 / 2 - 33 m g = 0 = change in total energy
so
v = sqrt (66 g)
= sqrt (66*9.81)
= 25.4 m/s
V^2 = Vo^2 + 2g*d
V^2 = 0 + 19.6*(58-25) = 646.8
V = 25.43 m/s.
To determine how fast the person is moving when they are 25.0m above the base of the hill, we can use the principle of conservation of energy.
When the person is at the top of the hill, they have gravitational potential energy equal to mgh, where m is their mass, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height of the hill (58.0m).
When the person is 25.0m above the base of the hill, their gravitational potential energy is mgh, where h is now 25.0m.
Since friction does not affect the person, there is no loss of mechanical energy, so the initial potential energy at the top is equal to the final potential energy at 25.0m above the base.
Therefore, we can set up the following equation:
mgh = mgh
Simplifying the equation, we have:
mgh - mgh = 0
Solving for h and g:
gh = gh
Since g is common on both sides, it cancels out:
h = h
Since the height h is the same on both sides of the equation, this means that the speed of the person is the same when they are 25.0m above the base as it was when they were at the top of the hill.
Therefore, the person's speed when they are 25.0m above the base of the hill is the same as the initial speed when they were at the top of the hill. In this case, we don't have enough information to determine the initial speed, so we cannot calculate the exact speed.