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  • 21-09-2017
  • Physics
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A ball is thrown in the air at an angle of 40∘. If the maximum height it reaches is 10m , what must be its initial speed?

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Аноним Аноним
  • 28-09-2017
Let v = the launch velocity.

Because the launch angle is 40° (with the horizontal), the initial vertical velocity is
v* sin(40°) = 0.6428v m/s.

Assume g = 9.8 m/s² and ignore air resistance.
At maximum height, the vertical velocity is zero.

Because the maximum vertical height is 10 m, therefore
(0.6428v m/s)² - 2*(9.8 m/s²)(10 m) = 0
0.4132v² = 196
v = 21.78 m/s

Answer: 21.8 m/s  (nearest tenth)
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