Asked by jcarrington
A bullet is fired from a gun at 300 m/s. It hits the ground at 3s later. At what angle (in degrees) above the horizon was the bullet fired? (Assume no air resistance)
Answers
Answered by
Elena
The time of the projectile motion is
t=2v₀sinα/g
sinα=gt/2v₀ =9.8•3/2•300=0.05
α=arcsin0.05=2.8°
t=2v₀sinα/g
sinα=gt/2v₀ =9.8•3/2•300=0.05
α=arcsin0.05=2.8°
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