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Help with M1 question

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    At time t = 0, two balls A and B are projected vertically upwards. The ball A is projected vertically upwards with speed 2m/s from a point 50m above the horizontal ground. The ball B is projected vertically upwards from the ground with speed 20m/s. At time t = T seconds, the two balls are at the same vertical height, h metres, above the ground. The balls are modelled as particles moving freely under gravity.

    I've got to the point where I have two equations:

    Ball A -
    h-50=2T+1/2(-9.8)t2

    Ball B -
    h=20T+1/2(-9.8)t2

    What do I have to do now to solve h and T?
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    Any help will be appreciated!
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    (Original post by ℓove)
    At time t = 0, two balls A and B are projected vertically upwards. The ball A is projected vertically upwards with speed 2m/s from a point 50m above the horizontal ground. The ball B is projected vertically upwards from the ground with speed 20m/s. At time t = T seconds, the two balls are at the same vertical height, h metres, above the ground. The balls are modelled as particles moving freely under gravity.

    I've got to the point where I have two equations:

    Ball A -
    h-50=2T+1/2(-9.8)t2

    Ball B -
    h=20T+1/2(-9.8)t2

    What do I have to do now to solve h and T?
    This is simply simultaneous equations, and so, assuming your equations are correct, we can subtract Ball A from Ball B, giving:

    h-(h-50)=20T+1/2(-9.8)T2 -(2T+ 1/2(-9.8)T2)

    50=18T

    T=50/18

    Then we can solve

    h=20*T+1/2(-9.8)T2 plugging in our T value

    Please note, I havent checked that your equations are correct.
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    (Original post by mathematigeek)
    This is simply simultaneous equations, and so, assuming your equations are correct, we can subtract Ball A from Ball B, giving:

    h-(h-50)=20T+1/2(-9.8)T2 -(2T+ 1/2(-9.8)T2)

    50=18T

    T=50/18

    Then we can solve

    h=20*T+1/2(-9.8)T2 plugging in our T value

    Please note, I havent checked that your equations are correct.
    Thanks!!!!!
 
 
 
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