The rope of a swing is 3.1 m long. Calculate the angle from the vertical at which a 77-kg man must begin to swing in order to have the same kinetic energy at the bottom as a 1430-kg car moving at 1.05 m/s (2.35 mi/hr).

KE=1/2 1430*1.05^2 solve that.

swing: PE at top=Ke at bottom
mgh=above KE
solve for h

Now, the angle.

draw the diagram. I see
cosTheta=(L-h)/L L is the swing rope length.
solve for Theta

To calculate the angle from the vertical at which the man must begin to swing, we need to find the gravitational potential energy and the kinetic energy at the bottom of the swing for both the man and the car.

First, let's calculate the kinetic energy of the car at the bottom of the swing:

Mass of the car, m1 = 1430 kg
Velocity of the car, v1 = 1.05 m/s

Kinetic energy of the car, KE1 = (1/2) * m1 * v1^2

Next, let's calculate the gravitational potential energy of the man at the bottom of the swing:

Mass of the man, m2 = 77 kg
Acceleration due to gravity, g = 9.8 m/s^2
Length of the rope, L = 3.1 m

Gravitational potential energy of the man, PE2 = m2 * g * L * (1 - cosθ)

Where θ is the angle from the vertical at the bottom of the swing.

Since the question mentions that the man must have the same kinetic energy as the car at the bottom of the swing, we can equate the kinetic energy of the car to the gravitational potential energy of the man:

KE1 = PE2

Substituting the values and rearranging the equation, we get:

(1/2) * m1 * v1^2 = m2 * g * L * (1 - cosθ)

Now, we can solve for θ:

θ = arccos(1 - [(1/2) * m1 * v1^2] / [m2 * g * L])

Substituting the given values, we get:

θ = arccos(1 - (1/2) * (1430 kg) * (1.05 m/s)^2 / (77 kg * 9.8 m/s^2 * 3.1 m)

Calculating this using a scientific calculator, we find:

θ ≈ 45.7 degrees

Therefore, the man must begin swinging at approximately 45.7 degrees from the vertical to have the same kinetic energy at the bottom as the car.