A cannon is fired from the top of the castle wall. The cannonball is fired at 50 m/s at an angle of 30 degrees. A cannonball that was accidentally dropped hits the moat below in 5s.

(a) How far from the castle wall does the cannonball hit the ground?

(b) What is the ball's maximum height above the ground?

Known: x(i) = t(i) = 0

v(i) = 50 m/s

v(ix) = (50m/s)(cos30) = 43m/s = v(fx)

v(iy) = (50m/s)(sin30) = 25m/s

a(x) = 0

a(y) = -9.8 m/s^2

y(f) = 0

For (a), I need to solve for x(f)

Step 1. Given t=1.5s for a cannonball to drop to the ground, I can calculate the height of the castle wall or y(i) = 1/2 (-9.8m/s^2) (1.5s)^2 = 11m

Step 2. Using the above value, solve for t:

y(f)=0=y(i) + v(iy)(t) + 1/2a(y)(t)^2

-11 = 25(t) - 4.9(t)^2

-11 = t[25 - 4.9t]

t = 0 or let 25 - 4.9t = 0

t = -25/-4.9 = 5.1s

Step 3, calculate x(f) = x(i) + v(ix)(t) + 1/2a(x)(t)^2

x(f) = 0 + 43(5.1) + 0 = 219 m

The answer in the text says it should be 239 m. Any insight on what I have done wrong?

How do I calculate (b)?

Thanks and sorry for the long post.