A cannon at ground level is firing at a fort 200m away with 20m high walls. It aims at an angle 30 degrees above the horizontal and fires cannonballs at 50m/s. Assuming no air resistance, will the cannonballs fall short, hit the walls or enter the fort?

solving horizontally, use v=s/t, where s=200, v=25(3)^1/2. Hence t=s/v=8/(3)^1/2 (where t is the time when the ball would reach the fort if it does not reach the ground)Solivng vertically, s=?, u=25, v is irrelevant, a=-9.8, t=8/(3)^1/2, as we worked out from resolving horizontally. (s is the height of the cannonball when it reaches the fort)Use s=ut+(at²)/2. s=25(8/(3)^1/2) + (-9.8)*(8/(3)^1/2)²/2s=11m (2sf). 0<11<20. Hence, when the cannonball reaches the fort it is above ground but below the top of the wall, so it hits the wall.