An antiaircraft gun fires at an elevation of 60° at an enemy aircraft at 10,000m above the ground. At what speed must the cannon be shot to hit the plane at that height.

Answer 1

Y^2 = Yo^2 + 2g*h = 0
Yo^2 = -2g*h = -2*(-9.8)*10,000 = 196,000
Yo = 443 m/s.=Ver. component of initial
velocity.
Vo = Yo/sin60 = 443/sin60 = 512 m/s =
Initial velocity = Speed at which the
cannon is fired.