c. We already know there's no acceleration from the x direction, so using the same conversions as in part a: xι = υιxο(t) + xιο
= υιxο(t)
Assuming the left cannon is to be at the origin. Now, like in part a, in order for there to be an impact, xr must also be equal to xι at some point, so just work out the math:
xr = υrxο(t) + xrο
= xι
υrxο(t) + xrο = υιxο(t)
xrο = υιxο(t) - υrxο(t)
t = xιο/(υιxο) - (υrxο)
= xιο/υι(cosΘ) - υr(cosφ)
It's very important to use the correct angles here.. The cannons are both pointed at each other, so Θis45° and φ has to be 120°
xrο = (1300)/100(cos45°) - 81.6(cos120°)
= 11.6 seconds
d. So now we know when the impact takes place, we just have to find where one of the projectiles was at the time. Calculating everything for the left projectile gives the coordinates relative to the left cannon, so:
y = (1/2)at² + υο(t) + yο
= -(1/2)gt² + υιyο(t)
= -(1/2)(9.8 m/s²)(11.6 s)² + (100sin45° m/s)(11.6 s)
= 160.9 m
x = υο(t)
= υιyο(t)
= (100cos45° m/s)(11.6 s)
= 820.2 m
Your coordinates are: x = 820.2 m, y = 160.9 m