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If a 0.08 kg cell phone falls off a table at 15 m/s, then what is its kinetic energy right before it hits the ground?

Sagot :

The kinetic energy of the phone right before it hits the ground is 9J.

Kinetic energy of the phone

The kinetic energy of the phone right before it hits the ground is calculated as follows;

K.E = ¹/₂mv²

where;

  • m is mass of the phone
  • v is velocity of the phone

K.E = ¹/₂(0.08)(15)²

K.E = 9 J

Thus, the kinetic energy of the phone right before it hits the ground is 9J.

Learn more about kinetic energy here: https://brainly.com/question/25959744

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