Distance traveled and hang time for 2D projectile

[Kstan333]

Homework Statement

A cannon is fired at an angle of 42 degrees, the cannon ball travels 8.9 meters in the x direction. Its hang time is 1.5 seconds. The cannon site .43 meters above the ground on a stand. If this same cannon was fired at 35 degrees, what would the hang time and the distance traveled be?

Homework Equations

The Attempt at a Solution

I tried to use velocity vectors and find the speed at which the cannon ball was fired, muzzle velocity, using A^2+B^2=C^2. Giving me a velocity of 8.85m/s, using this I attempted to use sine and cosine to find the velocitys in the x and y direction. I am stuck here.

 

[haruspex]

Must be something I'm not understanding. There seems to be too much information, and it conflicts. From the hang time (= time in the air, yes?) and the horizontal distance I can calculate horizontal speed. From that and the launch angle I can compute vertical speed at muzzle. From that, the hang time, and taking g=9.81 m/s² I compute that the cannonball lands 3.02 m below the level of the muzzle, but we're told the cannon is only .43 m above the ground.

 

[CWatters]

I agree...

Vx = 8.9/1.5 = 5.933 m/s
Then..
Vx = V Cos(42)
V = 7.984 m/s

Vy = V Sin(42)
= 5.342 m/s

Vertically
S=ut-0.5at²
S=5.342 * 1.5 - 0.5*9.8*1.5²

= 8.013 - 11.025
=-3m

So it does appear to land 3m below the cannon. So perhaps the ground isn't flat?

Perhaps an error in the height of the cannon?