Could I have some help with this suvat question?

A particle P is projected vertically upward from a point Owith speed 12ms - ¹ . One second after P has been projected from O , another particle is projected vertically upward from O with speed 20ms - ¹ . Find : a the time between the instant that P is projected from O and the instant when P and Q collide , b the distance of the point where P and Q collide from O?​
I've tried using s(t)=s(t+1) but it didn't work.

Im presuming P is time t and Q is time t+1? If so, think about the time shift for Q as when t=1, then Q must be at its initial conditions so its not t+1 its ...

Im presuming P is time t and Q is time t+1? If so, think about the time shift for Q as when t=1, then Q must be at its initial conditions so its not t+1 its ...

Oh so it's t-1 because P has been travelling longer?

I think an easier way to do this is not to try and relate P and Q's times to each other, but to keep them with the same time t and instead relate their initial displacements.

So if we focus on just P in the first second when it is projected, we should calculate how far it moves in this time:
s = ut + 0.5at^2
s = 20 x 1 + 0.5 x -9.8 x 1^2
s = 7.1

We should also calculate P's final velocity after one second:
v = u + at
v = 20 + -9.8 x 1
v = 2.2

Now, at the instant Q is projected (one second after P), we know P's initial displacement from O (7.1 metres) and its velocity (2.2 metres per second upwards). We can set Q's initial displacement from O to be 0 metres per second, and its velocity to be 20 metres per second upwards. From here, both particles are in the same reference frame in terms of time t, so we can now just equate them via simultaneous equations:

s_P = s_Q
2.2t +0.5 x -9.8 x t^2 + 7.1 = 20t + 0.5 x -9.8 x t^2
2.2t + 7.1 = 20t
t = 0.40 seconds (2.s.f.)

Then of course you have to add 1 second to the answer for the time it took Q to be projected in the first place.
So the final final solution is t = 1.40 seconds

Shown on Wolfram Alpha below:



Hope this helps! Let me know if you need help with part b) as well.

I think an easier way to do this is not to try and relate P and Q's times to each other, but to keep them with the same time t and instead relate their initial displacements.

So if we focus on just P in the first second when it is projected, we should calculate how far it moves in this time:
s = ut + 0.5at^2
s = 20 x 1 + 0.5 x -9.8 x 1^2
s = 7.1

We should also calculate P's final velocity after one second:
v = u + at
v = 20 + -9.8 x 1
v = 2.2

Now, at the instant Q is projected (one second after P), we know P's initial displacement from O (7.1 metres) and its velocity (2.2 metres per second upwards). We can set Q's initial displacement from O to be 0 metres per second, and its velocity to be 20 metres per second upwards. From here, both particles are in the same reference frame in terms of time t, so we can now just equate them via simultaneous equations:

s_P = s_Q
2.2t +0.5 x -9.8 x t^2 + 7.1 = 20t + 0.5 x -9.8 x t^2
2.2t + 7.1 = 20t
t = 0.40 seconds (2.s.f.)

Then of course you have to add 1 second to the answer for the time it took Q to be projected in the first place.
So the final final solution is t = 1.40 seconds

Shown on Wolfram Alpha below:



Hope this helps! Let me know if you need help with part b) as well.

Ok I understand. Thanks for the help

No prob, glad you found my method helpful!

If you don't mind, could you help me with this question:
A stone is dropped from the top of a building and 2 seconds later a stone is thrown vertically downwards with a speed 25ms^(-1). Both stones reach the ground at the same time. Find the height of the building.

Post what youve tried please

I tried s(t)=s(t-2) but it didn't work. I don't really understand this question tbh

What values of the variables did you choose?

Remember to be careful which direction you chose to be +

Stone 1 u = 0 etc

That should be correct? Assuming you're substituting t2 = t1 - 2 and not the other way around.

Your equation should be 4.9t1^2 = 25(t1-2) + 4.9(t1-2)^2
Then you get t1 (the time for stone 1 to reach the ground) as 5.63 seconds.

You can then plug that into another s = ut + 0.5at^2 equation (remembering to use the values for stone 2).
So s = 0 + 0.5 x 9.8 x 5.63^2 = 155 metres (3.s.f.)

But they're not starting from the same point so why do you equate them?

What do you mean sorry? Both are being dropped/thrown from the same point atop the building, so their values for s are the same. It doesn't matter that one event happened earlier/later than the other.

Unless you're referring to something else haha sorry if I misunderstood

Sorry I got confused whether they were dropped from the same point or not since it didn't say explicitly in the question. So do I just assume that they are?