Swansea Maths Exam prep, mechanics question HELP

Any chance anybody could help me with this? I've tried, but I haven't done M1,2 since last year

Cheers

Start by drawing a diagram and considering the SUVAT equations. Then look at the hint if you're still stuck.

the bottom bit. that's what I've got X to be equal to

I'm not quite sure what you've done or solved for. What are the equations that you started from?

I started with Xa and Ya, being the vertical and horizontal displacements of the attacker's cannonball. And Xd and Yd being the vertical and horizontal of the defender's cannonball.
r(n) is square root of n btw:

Xa=25t*r(3), Ya=25t-(gt^2)/2
Xd=100-25t, Yd=25t*r(3)-(gt^2)/2

I then rearranged each X for t, then substituted these into my Ys. This produced (with a bit of rearranging):

Ya= (1875x-2r(3)gx^2)/1875r(3)

Yd= [-2gx^2 + (400g-625r(3))x + (62500r(3) - 20000g)]/625

When they collide, Ya=Yd, therefore:

(1875x-2r(3)gx^2)/1875r(3) = [-2gx^2 + (400g-625r(3))x + (62500r(3) - 20000g)]/625

which I then multiplied out, and rearranged to get a quadratic:

4r(3)gx^2 + (7500-1200r(3)g)x + (60000r(3)g - 5625)=0

to then get me that discriminant

A few things.

1) Your equations for the defender's cannonball are not correct. The times at which each cannonball is launched are not the same. Remember, there is a delay after which the defender's cannonball is launched.

2) You are trying to find an expression that this delay time, T, satisfies; you are not trying to solve for x.

hmm, okay. So would my Xd = -25(t-T),

and my Yd = 25r(3)(t-T)-(g(t-T)^2)/2 ?

This then gets me to a simultaneous equation involving T and t? which i can then (hopefully) use to find the formula required?

Almost: Xd = 100 - 25(t-T), since we're defining horizontal displacement from the attacker. But yeah, you've got the right idea.