Solving Projectile Problem for Avalanche Angle

[veloix]

Homework Statement

A cannon with a muzzle speed of 992 m/s is used to start an avalanche on a mountain slope. The target is 1900 m from the cannon horizontally and 793 m above the cannon. At what angle, above the horizontal, should the cannon be fired? (Ignore air resistance.)

Homework Equations

Xf= xi+vit
Yf=yi+vit+1/2at2
vfy^2=viy^2+2a(yf-yi)
x=cos(theta)992m/s
Y=sin(theta)992m/s

The Attempt at a Solution

ok this one i can get all the equations setup.
xf= cos(theta)992m/s
793=(sin(theta)992m/s)t + 1/2(-9.80)t2
using the x eqaution i can solve for t.
t=1900/cos(theata)992m/s
but once i place it into the eqaution, it gets very messy and i get stuck on the arithmatic.
793= (sin(theta)992m/s)(1900/cos(theata)992m/s)+1/2(-9.80)(1900/cos(theata)992)^2
this is where i am stuck.

 

[Avodyne]

xf= cos(theta)992m/s

You have a position measured in meters per second. This obviously doesn't make sense.

 

[veloix]

You have a position measured in meters per second. This obviously doesn't make sense.

i know the it shuold look like this
xf=xi+vit
1900m= 0+(cos(theta)(992m/s)t