Question

If a body of mass 3 kg is dropped from a tower of the height of 25 m.
Then its kinetic energy at moment the body hits the ground is equal

Answer 1

Explanation:

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Answer 2

Answer:

490 J

Explanation:

As body is dropped, initial velocity is 0.

  Using v^2 = u^2 + 2aS: letters have their usual meanings.

⇒ v^2 = 0 + 2aS

⇒ v^2 = 2gS

We know, kinetic energy = (1/2)mv^2

⇒ 1/2 * m * v^2

⇒ 1/2 * 3 * 2gS    { v^2 = 2gS }

⇒ 2gS

⇒ 2 * g * 25 J

⇒ 50g J

  ⇒ 500 J( if g = 10 ), ⇒ 490 J(if g = 9.8 )