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    A ball A is thrown vertically downwards with speed 5ms^-1 from the top of a tower block 46m above the ground. At the same time as A is thrown downwards, another ball B is thrown vertically upwards from the ground with speed 18ms^-1. The balls collide.

    Find the distance of the point where A and B collide from the point where A was thrown.

    My working
    Ball A: u = 5, a = 9.8, s = ? (taking down +ve)
    Ball B: u = 18, a = -9.8, s = ? (taking down +ve)

    Ball A: s = 5t + 4.9t^2
    Ball B: s = 18t - 4.9t^2
    5t + 4.9t^2 = 18t - 4.9t^2
    9.8t^2 - 13t = 0
    t(9.8t - 13) = 0
    t = 13/9.8 = 1.32653...

    Substituting t into equation A...

    S = 5(13/9.8) + 4.9(13/9.8)^2
    S = 15.25m

    So it collides at 46m - 15.25m from the ground, since A was thrown off a tower 46m above ground. So the answer is 30.75m?

    My book however says the answer is 30m to 2.s.f., so have I done it wrong and fluked? or does the book just have the wrong answer?

    THANK YOU<3
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    Pretty sure the book is right. Your method is almost correct (equating two quadratics with t which are equal to s), but you missed something. You can't call both the distances 's', as they start 46 metres away from each other. Do you see what you should have done?

    Spoiler:
    Show
    [taking downwards as positive, as you did] if you call the distance ball a goes 's', ball b goes 's-46'
    Spoiler:
    Show
    move the 46 to the LHS so they both equal the same thing, then find t, and consequently s.


    Edit: Re-reading your method, I should add you should probably keep the accelerations the same direction, makes cancelling the t^2 easier, but don't believe it's essential.
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    (Original post by gcseeeman)
    Pretty sure the book is right. Your method is almost correct (equating two quadratics with t which are equal to s), but you missed something. You can't call both the distances 's', as they start 46 metres away from each other. Do you see what you should have done?

    Spoiler:
    Show
    [taking downwards as positive, as you did] if you call the distance ball a goes 's', ball b goes 's-46'
    Spoiler:
    Show
    move the 46 to the LHS so they both equal the same thing, then find t, and consequently s.


    Edit: Re-reading your method, I should add you should probably keep the accelerations the same direction, makes cancelling the t^2 easier, but don't believe it's essential.
    Using your way:
    -46 = 5t + 4.9t^2 (BALL A)
    0 = -18t + 4.9t^2 (BALL B)

    therefore 5t + 4.9t^2 + 46 = 4.9t^2 - 18t
    46 = -13t
    t = 3.53846...

    putting T into A:

    s = 5(46/13) + 4.9(46/13)^2
    s = 79.04378... umm :s
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    (Original post by dumb maths student)
    Using your way:
    -46 = 5t + 4.9t^2 (BALL A)
    0 = -18t + 4.9t^2 (BALL B)

    therefore 5t + 4.9t^2 + 46 = 4.9t^2 - 18t
    46 = -13t
    t = 3.53846...

    putting T into A:

    s = 5(46/13) + 4.9(46/13)^2
    s = 79.04378... umm :s
    Not exactly, you are mixing up the distances or something. I didn't check that thoroughly, but I think the minus 46 is on the wrong side. You also can't say one of them equals zero, it equals s. You just substitute it into the other one in place of s.
    As you have posted quite a bit of working, I'll show you what you should have done for t:
    Spoiler:
    Show

    For A
    s = s, t = t, a = 9.8, u = 5
    s = 5t + 4.9t^2

    For B
    s = (s-46) t = t, a = 9.8, u = -18
    s - 46 = -18 +4.9t^2 \rightarrow s = 46 -18t + 4.9t^2

    As they both equal s...
    5t + 4.9t^2 = 46 -18t + 4.9t^2
    23t = 46

    Can you see what to do now for the distance?
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    (Original post by gcseeeman)
    Not exactly, you are mixing up the distances or something. I didn't check that thoroughly, but I think the minus 46 is on the wrong side. You also can't say one of them equals zero, it equals s. You just substitute it into the other one in place of s.
    As you have posted quite a bit of working, I'll show you what you should have done for t:
    Spoiler:
    Show

    For A
    s = s, t = t, a = 9.8, u = 5
    s = 5t + 4.9t^2

    For B
    s = (s-46) t = t, a = 9.8, u = -18
    s - 46 = -18 +4.9t^2 \rightarrow s = 46 -18t + 4.9t^2

    As they both equal s...
    5t + 4.9t^2 = 46 -18t + 4.9t^2
    23t = 46

    Can you see what to do now for the distance?
    ah i see!

    thanks for the help i was sweating for like 30 minutes straight trying to do this haha
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    You're welcome
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    (Original post by dumb maths student)
    A ball A is thrown vertically downwards with speed 5ms^-1 from the top of a tower block 46m above the ground. At the same time as A is thrown downwards, another ball B is thrown vertically upwards from the ground with speed 18ms^-1. The balls collide.

    Find the distance of the point where A and B collide from the point where A was thrown.

    My working
    Ball A: u = 5, a = 9.8, s = ? (taking down +ve)
    Ball B: u = 18, a = -9.8, s = ? (taking down +ve)

    Ball A: s = 5t + 4.9t^2
    Ball B: s = 18t - 4.9t^2
    5t + 4.9t^2 = 18t - 4.9t^2
    9.8t^2 - 13t = 0
    t(9.8t - 13) = 0
    t = 13/9.8 = 1.32653...

    Substituting t into equation A...

    S = 5(13/9.8) + 4.9(13/9.8)^2
    S = 15.25m

    So it collides at 46m - 15.25m from the ground, since A was thrown off a tower 46m above ground. So the answer is 30.75m?

    My book however says the answer is 30m to 2.s.f., so have I done it wrong and fluked? or does the book just have the wrong answer?

    THANK YOU<3
    EDIT: Just realised that both values aren't s.

    Ball A is 46 - S
    Ball B is S

    EDIT2:

    Other way around ^^
 
 
 
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