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A 2.5 Kg watermelon is dropped off an 8-meter balcony. What will the watermelon’s velocity be right before it hits the ground?

Respuesta :

onyebuchinnaji

The watermelon’s velocity right before it hits the ground is 12.52 m/s.

Conservation of mechanical energy

The principle of conservation of mechanical energy states that the total energy of an isolated system is always conserved.

P.E = K.E

mgh = ¹/₂mv²

gh = ¹/₂v²

v = √2gh

where;

  • v is the velocity of the object before it hits the ground.
  • h is the height

The watermelon’s velocity right before it hits the ground is calculated as follows;

v = √(2 x 9.8 x 8)

v = 12.52 m/s

Learn more about conservation of mechanical energy here: https://brainly.com/question/6852965

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