A 2.00 kg cannonball is fired out of a cannon at an angle of 30.0 to the horizontal. When the cannonball reaches the top of its path, its momentum has a magnitude of 800 kg-m/s. What is the horizontal component of the cannonball's momentum when it left the cannon? Change in Momentum p=mΔv p=fnetΔT Impulse=Δp 3. Just learning momentum, but I don't know how to use the 30 degree angle with the momentum values to find the horizontal momentum? It does have to do with projectile motion, but I don't know how to use the data.