A-level Maths Mechanics: SUVAT Question HELP

Here is the question: A stone(1) is projected vertically upwards from a point O, with an initial speed of 8ms. At the same instant, another stone(2) is dropped from 5m above point O. Find the height above O at which the stones collide.

I have tried to model both stones by making the acceleration of both -9.8ms. Here is what I thought I knew about each stone.

Stone 1: Final s = s - 5, t=t, u=0, a = -9.8ms
Stone 2: s=s, t=t, u=8, t=-9.8ms

I am trying to find both displacements in terms of t and solve the equations simultaniously. Im stuck and don't know if I'm taking the right approach.

Sort of. Id sketch the diagram and mark on the positive direction (upwards you seem to choose), the origin O, the displacement above O (s), and the info for each stone. Then form the suvats, equate and solve.

Note you can simply solve it by thinking about the relative motion of the stones, but its worth getting the usual suvat approach sorted first.

Thanks

Upload your sketch and any working if you cant get it to work. In the OP you have u=8 for stone 2, and the initial height 5 seems to be applied to stone 1, so make sure its all consistent.

Heres the equations i formed:

for stone 1: 5-s = ut+1/2at^2 => s=4.9t^2 -5
for stone 2: s = ut+1/2at^2 => s=8t-4.9t^2

Is this correct?
What do I do next, set them as equal?

For the sketch, its vertical motion and Id draw it as such. Mark on postive upwards with an arrow.

S1 is at the bottom (the origin) and the displacement (positive upwards) represents the height above O which is what the question asks for . Its initial velocity is 8 (positive upwards). So s = ut + 1/2 at^2.

S2 is at the top (5m above (positive) the origin O). Its dispalcement above O can be modelled as s = 5 + ut + 1/2 at^2. Obv u=0 for this stone.

So then equate and solve.and when you get s, that represets the height above O which is what the question asks for. Your suvats etc may get something similar, but its worth getting it "correct" at the start.

Heres what I got:

5-4.9t^2 = 8t - 4.9t^2
5=8t
t=0.625

s = 3.09
so the height they meet is 3.09m

Does that seem right?

Seems about right. The relative motion approach says that as the stones have the same acceleration, Stone1 must approach Stone2 with a constant (zero relative acceleration) relative velocity of 8m/s. It must cover 5m for them to meet so t=5/8 and hence s = 5 - 4.9(5/8)^2.

So the suvats reduce to the above, as they should.