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kenny6666 [7]
2 years ago
13

A young diver is practicing his skills before an important team competition. Use the diagram below in order to analyze the energ

ies 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

Physics
2 answers:
kaheart [24]2 years ago
6 0
  • 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

skad [1K]2 years ago
5 0
  • 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

<u>Answers:</u>

<em><u>The </u></em><em><u>gravitational</u></em><em><u> potential</u></em><em><u> energy</u></em><em><u> of</u></em><em><u> the</u></em><em><u> </u></em><em><u>diver </u></em><em><u>when </u></em><em><u>he</u></em><em><u> reaches</u></em><em><u> point</u></em><em><u> C</u></em><em><u> </u></em><em><u>is </u></em><em><u>4</u></em><em><u>4</u></em><em><u>1</u></em><em><u>0</u></em><em><u> </u></em><em><u>J.</u></em>

<em><u>The </u></em><em><u>velocity</u></em><em><u> </u></em><em><u>of </u></em><em><u>the </u></em><em><u>diver </u></em><em><u>is </u></em><em><u>9</u></em><em><u>.</u></em><em><u>9</u></em><em><u> </u></em><em><u>m/</u></em><em><u>s.</u></em>

Hope you could get an idea from here.

Doubt clarification - use comment section.

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\huge \boxed { \sf{Answers}}

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