Asked by elle

A projectile is launched from the ground at 34.8 m/s at an angle of 44.5 degrees above the horizontal. How long, in seconds, does it take such that its angle of trajectory (it's velocity vector) is 20.6 degrees below the horizontal?

Answers

Answered by Steve
y(t) = 34.8 sin44.5° t - 4.9t^2
x(t) = 34.8 cos44.5° t

Now just find when

dy/dx = -tan20.6°

recall that dy/dx = (dy/dt) / (dx/dt)
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