Posted by duke on .
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?
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)