Question

A young diver is practicing his skills before an important team competition. Use the diagram below in order to analyze the energies of the diver and complete the statements below.

Where m = mass (kg), g = 9.8 m/s2, v = velocity (m/s), h = height (m), KE = kinetic energy (J), and GPE = gravitational potential energy (J).
Use the equations above to answer the following questions.

A diver with a mass of 90 kg is at a height of 10 m, and he has not jumped off of the board yet (v = 0 m/s). When the diver reaches a height of 5 m (Point C), his gravitational potential energy is

A. 1350 J

B. 8820 J

C. 4410 J

D. 0 J

and his velocity is

E. 4.5 m/s

F. 0 m/s

G. 3.2 m/s

H. 9.9 m/s

Please help will mark brainliest

Answer 1

• Mass of the diver (m) = 90 Kg.
• Height of the board from the ground (h) = 10 m.
• Acceleration due to gravity (g) = 9.8 m/s^2.
• Height of the diver from the ground when he reaches point C (x) = 5m
• Initial velocity (u) = 0 m/s
• We know, gravitational potential energy of a body = mass × acceleration due to gravity × height.
• Therefore, the gravitational potential energy of the diver when he reaches point C (GPE) = mg(h - x)
• or, GPE = [90 × 9.8 × (10-5)] J
• or, GPE = [90 × 9.8 × 5] J
• or, GPE = 4410 J
• For a freely falling body,
• v^2 - u^2 = 2gh
• or, v^2 = 2gh
• We know, kinetic energy of a body = 1/2 mv^2
• Therefore, kinetic energy of the diver when he reaches point C (KE) = 1/2 m(2gx)
• Here, 2gx = (2 × 9.8 × 5) = 98 (m/s)^2
• We have already seen v^2 = 2gh
• or, v = √2gh
• So, the velocity of the diver = √2gx = √98 m/s = 9.9 m/s

Answers:

The gravitational potential energy of the diver when he reaches point C is 4410 J.

The velocity of the diver is 9.9 m/s.

Hope you could get an idea from here.

Doubt clarification - use comment section.

Answer 2

• h=10m-5m=5m
• m=90Kg
• g=9.8m/s^2

$\\ \sf\longmapsto PE=mgh$

$\\ \sf\longmapsto PE=90(5)(9.8)$

$\\ \sf\longmapsto PE=4410J$

Now

It's converted to kinetic energy while reaching ground.

$\\ \sf\longmapsto K.E=4410$

$\\ \sf\longmapsto \dfrac{1}{2}mv^2=4410$

$\\ \sf\longmapsto 90v^2=8820$

$\\ \sf\longmapsto v^2=98$

$\\ \sf\longmapsto v=9.9m/s$