a cannon tilted up at 33 degree angle fires a cannon ball at 88 m/s from atop a 12.0 m high fortress wall. What’s the balls impact speed on the ground?

Answer 1

the height of the ball is
h(t) = 12 + (88 sin33°)t - 4.9t^2
h=0 at t=10
v(t) = 88 sin33° - 9.8t
so plug and chug

alternatively, consider the ball as being thrown down with an initial speed of
-88 sin33° m/s
and find how long it takes to drop 12m.
Plug that into the v(t) equation