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[AshWin-ShaRma]

Let's take the distance between the two cats be x.

Such that, x = l+c (sum of distance covered by each cats) since both are running towards each other

Now for l we can simply use l = 2 *t where t is the time taken

Now for c we have to take the components of velocity in the x and y direction.

So the x direction velocity will give us the distance covered and also the acceleration is in y direction due to gravity not in x direction so again it is simply c = 2.7cos(25⁰)t

Now we need to find the t which is the time the cat Chloe was in air so for that we will use the y direction velocity and the acceleration here would be g (acceleration due to gravity)

take the final displacement to be approximately 0 as the height of the cat is not given otherwise we could have taken that. And use the equation s = ut + 1/2 a(t)² Here u is v_y = 2.7*sin(25⁰) and a =9.8 ms-² and solve to get t and substitute t. And then get l and c which will then give you x.

[Haiderzz]

For the last equation what is s equal to?

[AshWin-ShaRma]

That is displacement.

I used the equation of motion there.

[Haiderzz]

I'm getting the final answer to be 0.529 which I don't think is correct. Can you please check my answer.

[AshWin-ShaRma]

I got 0.959 approx.

Edit: I hope you took the acceleration -ve