Two cannons, one aimed 45° (left) from horizontal, the other aimed 60° (right) from horizontal, are..

simultaneously fired toward each other.

a.
In order for the two projectiles to collide, at some point in time the y components of position must be the same. Since the cannons were fired simultaneously, show that this means the y components of their initial velocities must also be equal.

b.
Given what you showed in part a, and that projectiles are fired from the left cannon at 100 m/s, determine the required initial velocity of projectiles fired out of the right cannon for an impact to occur.

c.
If the cannons are 1.3 km apart, when does this impact take place?

d.
Where does it take place? (Give x,y coordinates relative to the left-hand cannon)

Remember that in order for things to collide, they must be in the same position at the same time.

a. Let's just make a few definitions first, to start us off:

yι = y component of left projectile's displacement
yιο = y component of left projectile's initial position
yr = y component of right projectile's displacement
yrο = y component of right projectile's initial position
υιyο =y component of left projectile's initial velocity
υryο = y component of right projectile's initial velocity
t = time

yι = (1/2)at² + υιyο(t) + yιο
= υιyο(t) - (1/2)gt²

yr = (1/2)at² + υryο(t) + yrο
=υryο(t) - (1/2)gt²

For an impact to happen, yι and yr have to be equal at some time:
yι = yr
υιyο(t) - (1/2)gt² = υryο(t) - (1/2)gt²
υιyο(t) = υryο(t)
υιyο = υryο

Time cancels out of the equations, so the initial velocities are equal and so are the y components of the positions of the projectiles. Both are being accelerated by the same force, and have the same initial conditions, so their motion in the y direction will be exactly the same.

b. You just play with your vectors here:

υιyο = υryο
υιο(sinΘ) = υrο(sinΘ)
υιο = υrο(sinΘ)/(sinΘ)
= 100(sin45°)/(sin60°)