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 18.0 m/s via path 1. How fast is he moving on path 2 when he releases the rope at a height of 4.35 m above the water? Ignore the effects of air resistance.
To determine the speed at which the person is moving on path 2 when releasing the rope, we can use the principle of conservation of mechanical energy. The total mechanical energy of the person, consisting of potential energy and kinetic energy, remains constant throughout the motion.
Let's define the following variables:
- v1: speed of the person on path 1 when releasing the rope (given: 18.0 m/s)
- h1: height of the cliff on path 1
- h2: height of the cliff on path 2 (given: 4.35 m)
- v2: speed of the person on path 2 when releasing the rope (unknown)
The conservation of mechanical energy equation can be written as follows:
mgh1 + (1/2)mv1^2 = mgh2 + (1/2)mv2^2
Since the mass (m) of the person cancels out on both sides, we can simplify the equation to:
gh1 + (1/2)v1^2 = gh2 + (1/2)v2^2
We can rearrange the equation and solve for v2:
v2^2 = (2gh1 + v1^2 - 2gh2)
Now, substitute the known values into the equation:
v2^2 = (2 * 9.8 m/s^2 * h1 + (18.0 m/s)^2 - 2 * 9.8 m/s^2 * 4.35 m)
Calculate the value inside the parentheses:
v2^2 = (19.6 h1 + 324 - 85.26)
Therefore,
v2^2 = (19.6 h1 + 238.74)
Finally, take the square root of both sides to find the speed on path 2:
v2 = √(19.6 h1 + 238.74)
Note: To find the value of v2, you need to specify the height of the cliff on path 1 (h1).