a person is standing at the top of a 58 m tall hill. He stumbles and falls. As he rolls down the hill, magically friction has no effect on him! when he is still 25.0 m above the base of the hill determine how fast he is moving.
V^2 = Vo^2 + 2g*d.
V^2 = 0 + 19.8*(58-25) = 646.8
V = 25.4 m/s.
To determine how fast the person is moving when he is 25.0 m above the base of the hill, we can use the principle of conservation of energy.
The total mechanical energy of the person is conserved as he rolls down the hill, neglecting the effect of friction. At the top of the hill, the person only has potential energy due to his height above the base. As he rolls down the hill, this potential energy is converted into kinetic energy.
Using the conservation of energy principle, we equate the initial potential energy at the top of the hill to the final kinetic energy when the person is 25.0 m above the base of the hill.
Initial potential energy (UP) = Final kinetic energy (KE)
The initial potential energy is given by the formula:
UP = m * g * h
Where:
m is the mass of the person (which we will assume to be 1 kg)
g is the acceleration due to gravity (approximately 9.8 m/s^2)
h is the height of the hill (58 m)
Substituting these values into the equation, we get:
UP = (1 kg) * (9.8 m/s^2) * (58 m)
Next, we find the velocity at 25.0 m above the base of the hill using the formula for kinetic energy:
KE = (1/2) * m * v^2
Where:
v is the velocity of the person
Rearranging the equation and substituting the known values, we get:
v = sqrt((2 * KE) / m)
To find KE, we can use the conservation of energy equation from earlier:
KE = UP = (1 kg) * (9.8 m/s^2) * (58 m)
Substituting this into the equation for v, we get:
v = sqrt((2 * (1 kg) * (9.8 m/s^2) * (58 m)) / (1 kg))
Simplifying this expression and calculating the value of v will give us the speed at that point.