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# Mechanics suvat question watch

1. I'm in year 13 and have decided to take up AS Further Maths. I'm taking Physics, so I am very familiar with the suvat equations, however I am stuck on a question. I've done very similar questions before, but I can't remember how. Here it is:

"A ball A is thrown vertically upwards at 25m/s from a point P. Three seconds later a second ball B is also thrown vertically upwards from point P at 25m/s. g=10m/s^2.
Calculate: (i) the time for which ball A has been in motion when the balls meet, (ii) the height above P at which A and B meet."

Am I correct in thinking that the balls would collide as A was falling and B was still moving upwards?
Here's what I've done so far:

For ball A: v=0, u=25, a=-10
t=(v-u)/a = 2.5s
So ball A it at max. height after 2.5 seconds. That means it is falling for 0.5 seconds before ball B is released.

v^2=u^2-2as
25^2=20s
s=31.25m at the top.

When ball B is released, at t=3:
s=5=0.5^2 = 1.25
31.25-1.25=30m
So ball A is 30m above point P.
Its velocity at that point is 5m/s (using v=u+at).

This is the part where I get stuck. Every time I try to visualise it, my mind just blanks completely. I know that this is going to involve simultaneous equations, but I've found so many pairs of equations and none of them work.

I can handle part (ii), it's just that I need part (i) first.

2. (Original post by JustJusty)
I'm in year 13 and have decided to take up AS Further Maths. I'm taking Physics, so I am very familiar with the suvat equations, however I am stuck on a question. I've done very similar questions before, but I can't remember how. Here it is:

"A ball A is thrown vertically upwards at 25m/s from a point P. Three seconds later a second ball B is also thrown vertically upwards from point P at 25m/s. g=10m/s^2.
Calculate: (i) the time for which ball A has been in motion when the balls meet, (ii) the height above P at which A and B meet."

Am I correct in thinking that the balls would collide as A was falling and B was still moving upwards?
Here's what I've done so far:

For ball A: v=0, u=25, a=-10
t=(v-u)/a = 2.5s
So ball A it at max. height after 2.5 seconds. That means it is falling for 0.5 seconds before ball B is released.

v^2=u^2-2as
25^2=20s
s=31.25m at the top.

When ball B is released, at t=3:
s=5=0.5^2 = 1.25
31.25-1.25=30m
So ball A is 30m above point P.
Its velocity at that point is 5m/s (using v=u+at).

This is the part where I get stuck. Every time I try to visualise it, my mind just blanks completely. I know that this is going to involve simultaneous equations, but I've found so many pairs of equations and none of them work.

I can handle part (ii), it's just that I need part (i) first.

s u a and t are most common in forming simultaneous equations

time is the thing which is slightly confusing

you will have 2 unknows so don't worry that's why u gonna make a simultaneous equations

let t be the time at which ball A is thrown up so if t for A is just t

then the time for B i=will be t+3 since after t seconds B waits another 3 before being thrown
3. (Original post by will'o'wisp)
s u a and t are most common in forming simultaneous equations

time is the thing which is slightly confusing

you will have 2 unknows so don't worry that's why u gonna make a simultaneous equations

let t be the time at which ball A is thrown up so if t for A is just t

then the time for B i=will be t+3 since after t seconds B waits another 3 before being thrown

For ball A: a=10, u=0
s=0t+5t^2

For ball B: a=-10, u=25
s=25t-5t^2

5t^2 = 25t - 5t^2
10t = 25
t = 2.5

t here is the time from the point where ball A started falling.
The displacement from that point is:
s=0+(5x2.5^2)
s=31.25

This is the maximum displacement, so if this is true, they collide exactly at point P, which makes no sense since ball A was 30 metres in the air at the point when ball B was released.

What am I doing wrong?
4. (Original post by JustJusty)

For ball A: a=10, u=0
s=0t+5t^2

For ball B: a=-10, u=25
s=25t-5t^2

5t^2 = 25t - 5t^2
10t = 25
t = 2.5

t here is the time from the point where ball A started falling.
The displacement from that point is:
s=0+(5x2.5^2)
s=31.25

This is the maximum displacement, so if this is true, they collide exactly at point P, which makes no sense since ball A was 30 metres in the air at the point when ball B was released.

What am I doing wrong?
Ok, personally i think you're missing out a few things here

defining up as positive
for A u=25 t=t a=-10 s=s

for B u=25 t=t+3 a=-10 s=s

i'll let you go from there, ask if you don't understand what i've done
5. (Original post by will'o'wisp)
Ok, personally i think you're missing out a few things here

defining up as positive
for A u=25 t=t a=-10 s=s

for B u=25 t=t+3 a=-10 s=s

i'll let you go from there, ask if you don't understand what i've done
Okay, this makes sense.
I got t=-4 which is a plausible answer (if I ignore the negative).

Thanks!

**edit: never mind, I forgot to make some things negative before and now I got t=1 :/
6. (Original post by JustJusty)
Okay, this makes sense.
I got t=-4 which is a plausible answer (if I ignore the negative).

Thanks!

**edit: never mind, I forgot to make some things negative before and now I got t=1 :/
good stuff
7. (Original post by will'o'wisp)
s u a and t are most common in forming simultaneous equations

time is the thing which is slightly confusing

you will have 2 unknows so don't worry that's why u gonna make a simultaneous equations

let t be the time at which ball A is thrown up so if t for A is just t

then the time for B i=will be t+3 since after t seconds B waits another 3 before being thrown
(Original post by JustJusty)
Okay, this makes sense.
I got t=-4 which is a plausible answer (if I ignore the negative).

Thanks!

**edit: never mind, I forgot to make some things negative before and now I got t=1 :/
B's time is t - 3 since it is travelling for a SMALLER time than A.
8. (Original post by tiny hobbit)
B's time is t - 3 since it is travelling for a SMALLER time than A.
There, whoops my mistake :/ I knew it went wrong somewhere >.> Thanks for clearing it up

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