Finding when two particles fired at different times reach the same height

[Taylor_1989]

I would just like someone to give a quick check to part b) of this question, just to make sure I have fully grasped the concept.

A t=0s, particle P is fired vertically upwards with speed 17.64ms^-1.When P reaches its maximum height, particle Q is fired vertically upwards with a speed 25ms^-1. Both particles are fired from ground level.

a) find the time between the two particles being fired.
Ans:1.8s

b) Find the length of the time after P is fired before the two particles become level.

So I start with the SUVAT equation:

S=S_p, U=17.64, V=0, T=T
S=S_q, U=25, V=N/A, T=T-1.8

I then use the equation $S=ut+\frac{1}{2}at^2$

I would the set S_p=S_q, and solve simultaneously.

My point I would like to touch on is the t part. I set T=T because I do not know the total time take for the ball to reach maximum height; correct? T=T-1.8, because once again I do not know 1. the time taken for the ball to reach max height 2. - 1.8 because it was fired 1.8s later than the first.

I did end up with the right ans which is t=2.44s. But just want to makes sure I have the right intuition behind the question.

Big thanks in advance.

 

[Doc Al]

I did end up with the right ans which is t=2.44s. But just want to makes sure I have the right intuition behind the question.

Yes you do. Good job.