A 56 kg diver runs and dives from the edge of a cliff into the water which is located 4.0 m below. If she is moving at 8.0 m/s the instant she leaves the cliff, determine the following. Her gravitational potential energy relative to the water surface when she leaves the cliff
Her kinetic energy when she leaves the cliff
Her total mechanical energy relative to the water surface when she leaves the cliff
Her total mechanical energy relative to the water surface just before she enters the water.
The speed at which she enters the water.

1) Her gravitational potential energy relative to the water surface when she leaves the cliff is; GPE(leaves cliff) = 2195.2 J

2) Her kinetic energy when she leaves the cliff is; KE = 0J

3) Her total mechanical energy relative to the water surface when she leaves the cliff is; ME_total = 2195.2 J

4) Her total mechanical energy relative to the water surface just before she enters the water is; ME_total = 2195.2 J

5) The speed at which she enters the water is; v = 8.85 m/s

We are given;

Mass of the diver; m = 56 kg

Height of the cliff; h = 4 m

Speed at which she is moving; vₓ = 8 m/s

1) Formula for gravitational potential energy is;

GPE = mgh

where;

m is mass

g is acceleration due to gravity

h is height

Thus;

GPE = 56 × 4 × 9.8

GPE(leaves cliff) = 2195.2 J

2) The formula for kinetic energy when she leaves the cliff is;

KE = ¹/₂mu²

Where;

m is mass

u = initial velocity = 0 m/s

Thus;

KE = ¹/₂ × 56 × 0²

KE(leaves cliff) = 0 J

3) The total mechanical energy relative to the water surface when she leaves the cliffis;

ME_total = GPE(leaves cliff) + KE(leaves cliff)

Thus;

ME_total = 2195.2 + 0

ME_total = 2195.2 J  

4) Her total mechanical energy relative to the water surface just before she enters the water is same as that when she leaves the cliff = 2195.2 J

5) The speed with which she enters the water, v, is gotten from newtons third equation of motion;

v² = u² + 2gh

Thus;

v² = 0² + (2 × 9.8 × 4)

v² = 78.4

v = √78.4

v = 8.85 m/s

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1) 2197.44 J

2) 0 J

3) 2197.44 J = Constant

4) 2197.44 J

5) Approximately 8.86 m/s

Explanation:

The given parameters are;

The mass of the diver, m = 56 kg

The height of the cliff, h = 4.0 m

The speed with which the diver is moving, vₓ = 8.0 m/s

The gravitational potential energy = Mass, m × Height of the cliff, h × Acceleration due to gravity, g

1) Her gravitational potential energy = 56 × 4.0 × 9.81 = 2197.44 J

2) The kinetic energy = 1/2·m·u²

Where;

u = Her initial velocity = 0 when she just leaves the cliff

Therefore;

Her kinetic energy when she just leaves the cliff = 1/2 × 56 × 0² = 0 J

3) The total mechanical energy = Kinetic energy + Potential energy

The total mechanical energy is constant

Her total mechanical energy relative to the water surface when she leaves the cliff = Her gravitational potential energy = 2197.44 J = Constant

4) Her total mechanical energy relative to the water surface just before she enters the water = 2197.44 J

5) The speed with which she enters the water, v, is given from, v² = u² + 2·g·h

Where;

u = The initial velocity at the top of the cliff before she jumps= 0 m/s

∴ v² = 0² + 2 × 9.81 × 4 = 78.48

v = √78.48 ≈ 8.86 m/s

The speed with which she enters the water, v ≈ 8.86 m/s

jupiter is the largest planet in our solar system, having a mass and radius that are respectively, 318 and 11.2 times that of earth. suppose that an object falls from rest near the surface of each planet and that the acceleration due to gravity remains constant during the fall. each object falls the same distance before striking the ground. determine the ratio of the time of fall on jupiter to that on earth.