The drawing below shows a person who, starting from rest at the top of a cliff, swings down at the end of a rope, releases it, and falls into the water below. There are two paths by which the person can enter the water. Suppose he enters the water at a speed of 18.5 m/s via path 1. How fast is he moving on path 2 when he releases the rope at a height of 4.00 m above the water? Ignore the effects of air resistance.

To determine the velocity of the person on path 2 when they release the rope, we can use the principle of conservation of mechanical energy. The person starts at rest on path 1, so their initial kinetic energy is zero. When they reach the water, their potential energy is also zero. Therefore, we can equate the initial potential energy on path 1 to the final kinetic energy on path 2.

The initial potential energy can be calculated using the formula:

PE = mgh

Where:
PE = Potential energy
m = Mass of the person (which we can assume to be constant)
g = Acceleration due to gravity (approximately 9.8 m/s²)
h = Height of the cliff (given as 4.00 m)

So, the initial potential energy on path 1 is equal to mgh.

Next, we find the final kinetic energy on path 2. We know that the person enters the water with a speed of 18.5 m/s on path 1. The formula for kinetic energy is:

KE = (1/2)mv²

Where:
KE = Kinetic energy
m = Mass of the person (constant as assumed)
v = Final velocity on path 2 (to be determined)

Therefore, the final kinetic energy on path 2 is equal to (1/2)mv².

Now, based on the conservation of mechanical energy, we can equate the initial potential energy on path 1 to the final kinetic energy on path 2:

mgh = (1/2)mv²

Mass (m) is common to both sides of the equation, so it cancels out:

gh = (1/2)v²

To find the velocity on path 2 (v), we rearrange the equation:

v² = 2gh

Finally, taking the square root of both sides, we get:

v = √(2gh)

Plugging in the values for g (approximately 9.8 m/s²) and h (4.00 m), we can calculate the velocity:

v = √(2 * 9.8 * 4.00)

v ≈ 8.85 m/s

Therefore, the person's velocity on path 2 when they release the rope at a height of 4.00 m above the water is approximately 8.85 m/s.