A cannon is fired from a cliff 190 m high downward at an angle of 38o with respect to the horizontal. If the muzzle velocity is 41 m/s, what is its speed (in m/s) when it hits the ground?
Which equations do I need for this?
V^2 = sqrt[Vx^2 + Vy^2]
Vx remains 41 m/s * cos 38 during flight.
Vy at impact can be calculated using
Vy^2 = (V sin 38)^2 + 2 g H
V is 41 m/s, right?
I keep getting 64.28 m/s as my answer, but it's not right. What am I doing wrong?