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A ball is tossed upwards into the air with an initial upwards velocity of 3.0m/s. What is the ball's velocity when it returns to the height from which it was tossed?

Sagot :

  • Initial velocity=u=3.0m/s
  • Acceleration due to gravity=g=9.8m/s^2

Neglate air resistance.

  • So acceleration is unchanged
  • As acceleration is remaining constant so at the top the velocity is 0.
  • But at the end the velocity again come to 3m/s.
Answer :

Explanation : h =v²/2g
h =9m²/s²/20m/s²
h = 0.45 m
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