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One of the Lady Spartans was falling to the ground after
dunking the winning basket. At the end of her fall, she was
falling at 4 m/s. If she was 60 kg, how much potential energy
did she have at the top of her jump?

Respuesta :

Kazeemsodikisola

Answer:

Explanation:

Given that,

A lady falling has a final velocity of 4m/s

v = 4m/s

Mass of the lady is 60kg.

m = 60kg

Using conservation of energy, the potential energy of the body from the point where the lady is dropping is converted to the final kinetic energy of the lady.

Therefore,

P.E = K.E(final) = ½mv²

P.E = ½ × 60 × 4²

P.E = 480 J.

nuhulawal20

At the top of the Lady Spartan's jump, she had a potential energy of 480J.

Given the data in the question;

  • Velocity of the Lady spartan;[tex]v = 4m/s[/tex]
  • Mass of the Lady spartan; [tex]m = 60kg[/tex]

Using the Conservation Energy:

When the lady spartan began falling, her potential energy becomes her kinetic energy.

Potential Energy = Kinetic Energy

[tex]P.E = K.E[/tex]

And [tex]K.E = \frac{1}{2}mv^2[/tex]

Hence: [tex]P.E = \frac{1}{2}mv^2[/tex]

We substitute our given values into the equation

[tex]P.E = \frac{1}{2} * 60kg * (4m/s)^2\\\\P.E = 480 kg.m^2/s^2\\\\ P.E = 480 J[/tex]

Therefore, at the top of the Lady Spartan's jump, she had a potential energy of 480J.

Learn more: https://brainly.com/question/999862

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