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?

Answer 1

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)