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recoiling problem help :(
I've included a picture also.
A circus cannon, which has a mass M = 4000 kg, is tilted at theta = 40°. When it shoots a projectile at v0 = 90 m/s with respect to the cannon, the cannon recoils along a horizontal track at vcannon = 1 m/s with respect to the ground.

a) At what angle to the horizontal does the projectile move with respect to the ground?
b) What is the mass of the projectile?
c) The cannon is now lowered to shoot horizontally. It fires the same projectile at the same speed relative to the cannon. With what speed does the cannon now recoil with respect to the ground?
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a) At first I thought the angle with respect to the ground is just 40 degrees, but it isn't. I still don't know how to do this, but my buddy figured out that the answer is 40*sin (1)+40=, but even he doesn't know why it works. Can somebody show me how to do this correctly?
b) Well, using the conservation of momentum, momentum of cannon=momentum of the projectile
so
4000kg(1m/s)=m(90m/s*cos(40.7)
m= 58.6 kg
c) I'm stuck. I figured that since the cannon is lowered to shoot horizontally, the angle would be 0, so I tried to equate the momentum of the cannon with the projectile to get:
4000kg*V=(58.6kg)(90 m/s*cos (0))
to get V=1.32m/s, but its not working out for some reason. Whats up with that?
I've included a picture also.
A circus cannon, which has a mass M = 4000 kg, is tilted at theta = 40°. When it shoots a projectile at v0 = 90 m/s with respect to the cannon, the cannon recoils along a horizontal track at vcannon = 1 m/s with respect to the ground.

a) At what angle to the horizontal does the projectile move with respect to the ground?
b) What is the mass of the projectile?
c) The cannon is now lowered to shoot horizontally. It fires the same projectile at the same speed relative to the cannon. With what speed does the cannon now recoil with respect to the ground?
================
a) At first I thought the angle with respect to the ground is just 40 degrees, but it isn't. I still don't know how to do this, but my buddy figured out that the answer is 40*sin (1)+40=, but even he doesn't know why it works. Can somebody show me how to do this correctly?
b) Well, using the conservation of momentum, momentum of cannon=momentum of the projectile
so
4000kg(1m/s)=m(90m/s*cos(40.7)
m= 58.6 kg
c) I'm stuck. I figured that since the cannon is lowered to shoot horizontally, the angle would be 0, so I tried to equate the momentum of the cannon with the projectile to get:
4000kg*V=(58.6kg)(90 m/s*cos (0))
to get V=1.32m/s, but its not working out for some reason. Whats up with that?
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