Posted by tibby on .
A cannon with a muzzle speed of 1008 m/s is used to start an avalanche on a mountain slope. The target is 1900 m from the cannon horizontally and 802 m above the cannon. At what angle, above the horizontal, should the cannon be fired? (Ignore air resistance.)
Someone help please. I've tried two different approaches to this problem which were done by other students also and it worked for the same question but with different numbers. I've used this formula
x= horizontal distance of cannon from target y= meters above the cannon
g= 9.8 u= cannon's muzzle speed
r= unknown a= angle at which cannon should be fired
b= theta

r = sqrt(x^2+y^2)
cos(b)= x/r
r*sin(2ab)=(9.8*2^2)+y
i got 62.429 degree
and the second method used was this formula
h=(v^2sin^2theta)/2(9.8)
and i've gotten 11.643 degree
both of them are wrong, please help :(
Due tonight at 12am

physics 
tibby,
nvm there was a arithmetic error