A cannon tilted up at a 26.0 degree angle fires a cannon ball at 78.0 m/s from atop a 12.0 m-high fortress wall. What is the ball's impact speed on the ground below?

Answer 1

u = constant horizontal speed = 78 cos 26

Vi = initial up speed = 78 sin 26

v = Vi - g t

h = 12 + Vi t - 4.9 t^2

at ground, h = 0
4.9 t^2 -Vi t - 12 = 0

solve for t, time in air
then use that t so solve for v at t in v=Vi -9.8 t

then impact speed = sqrt(u^2+v^2)