Turn on thread page Beta
    • Thread Starter
    Offline

    15
    ReputationRep:
    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.

    Thanks in advance!
    Offline

    3
    ReputationRep:
    (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.

    Thanks in advance!
    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
    • Thread Starter
    Offline

    15
    ReputationRep:
    (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
    I've already done that but the answer doesn't make sense.

    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?
    Offline

    3
    ReputationRep:
    (Original post by JustJusty)
    I've already done that but the answer doesn't make sense.

    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
    s=ut+\frac{1}{2} at^2
    s=25t-5t^2

    for B u=25 t=t+3 a=-10 s=s
    s=ut+\frac{1}{2} at^2
    s=25(t+3) -5(t+3)^2

    i'll let you go from there, ask if you don't understand what i've done
    • Thread Starter
    Offline

    15
    ReputationRep:
    (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
    s=ut+\frac{1}{2} at^2
    s=25t-5t^2

    for B u=25 t=t+3 a=-10 s=s
    s=ut+\frac{1}{2} at^2
    s=25(t+3) -5(t+3)^2

    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 :/
    Offline

    3
    ReputationRep:
    (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
    Offline

    15
    ReputationRep:
    (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.
    Offline

    3
    ReputationRep:
    (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
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: November 6, 2016
The home of Results and Clearing

2,039

people online now

1,567,000

students helped last year

University open days

  1. London Metropolitan University
    Undergraduate Open Day Undergraduate
    Sat, 18 Aug '18
  2. Edge Hill University
    All Faculties Undergraduate
    Sat, 18 Aug '18
  3. Bournemouth University
    Clearing Open Day Undergraduate
    Sat, 18 Aug '18
Poll
A-level students - how do you feel about your results?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.