The cable of a 1800-kg elevator has broken, and the elevator is moving downward at a steady speed of 1.6 m/s. A safety braking system that works on friction prevents the downward speed from increasing.

a. speed of boat when leaves ramp (assume that engine turned off once boat hits ramp)

b. what is the speed of the boat when it hits the water again. (neglect air resistance)

To determine the speed of the boat when it leaves the ramp and when it hits the water again, we can analyze the given information using the principles of conservation of energy and motion.

a. To find the speed of the boat when it leaves the ramp, we need to determine the potential energy of the boat at that point. The potential energy can be calculated using the formula:

Potential energy = mass * gravity * height

Given:
Mass of the elevator (m) = 1800 kg
Height of the ramp (h) = ?

We need to first find the height of the ramp. Since the elevator is moving downward at a steady speed of 1.6 m/s, we can assume that the gravitational potential energy is being converted to kinetic energy and the elevator is losing potential energy at a constant rate. Therefore, we can equate the potential energy to the kinetic energy.

Potential energy = Kinetic energy
mgh = (1/2)mv^2

Simplifying the equation, we get:
gh = (1/2)v^2

Substituting the known values,
9.8 * h = (1/2) * (1.6)^2

Solving for h:
h ≈ 0.131 m

Therefore, the height of the ramp is approximately 0.131 meters.

b. To find the speed of the boat when it hits the water again, we need to consider the conservation of mechanical energy. Since the boat loses potential energy while descending the ramp, it gains an equal amount of kinetic energy.

Using the equation:
Potential energy = Kinetic energy
mgh = (1/2)mv^2

We can solve for v, the velocity of the boat when it hits the water.

gh = (1/2)v^2
9.8 * 0.131 = (1/2) * v^2

Solving for v:
v ≈ 1.27 m/s

Therefore, the speed of the boat when it hits the water again is approximately 1.27 m/s.