Question
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
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)
x(t) = 34.8 cos44.5° t
Now just find when
dy/dx = -tan20.6°
recall that dy/dx = (dy/dt) / (dx/dt)