# 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.
Original post by BlackLives
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 ...
(edited 2 months ago)
Original post by mqb2766
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 ...

Or yes you could do s(t) = s(t-1), it's kind of a trick question because you expect it to be t+1 but it's the opposite
Original post by mqb2766
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?
Original post by BlackLives
Oh so it's t-1 because P has been travelling longer?

Yes. When Ps time is 1, then the time for Q is 0 so t-1.
Original post by BlueBazooka
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.

Please remove the details of your working - it is against form rules to post full solutions, Thanks.
Original post by BlueBazooka
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 🙂.
(edited 2 months ago)
Original post by BlackLives
Ok I understand. Thanks for the help

Original post by BlueBazooka

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.
Original post by BlackLives
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.

Original post by Muttley79

I tried s(t)=s(t-2) but it didn't work. I don't really understand this question tbh
Original post by BlackLives
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
Original post by Muttley79
What values of the variables did you choose?

Remember to be careful which direction you chose to be +

Stone 1 u = 0 etc

I did u= 0 a=9.8 t=t for stone 1 and u=25 a=9.8 t= t-2 for stone 2.
Original post by BlueBazooka
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?
Original post by BlackLives
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
Original post by Moreover
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?
Original post by BlackLives
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?

Yeah I think you're supposed to assume that they're thrown from the same point, or else it'd be an impossible question haha