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M1 June 2013 (R) Question 4 help Watch

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    Video's on this paper are impossible to find (idk why). Im completely struggling with Question 4 especially. The mark scheme 100% confuses me. Can someone help explain how to do Question 4 for the M1 June 2013 (R)? Much appreciated.

    Here's the question:
    At time t = 0, two balls A and B are projected vertically upwards. The ball A is projectedvertically upwards with speed 2 m s–1 from a point 50 m above the horizontal ground.The ball B is projected vertically upwards from the ground with speed 20 m s–1. At timet = 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. Find (a) the value of T, (5)
    (b) the value of h.(2)
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    What have you tried so far?
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    (Original post by samb1234)
    What have you tried so far?
    Well i might sound a little stupid but I drew a diagram of the motion of the two particles. Then i came up with the equation. using s=ut + 0.5at^2

    50 + h = 2T + 0.5 x -9.8 x T^2
    and
    50= 20T + 0.5 x -9.8 x T^2

    I don't know if im doing alright so far and i have no clue where to go from here
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    Well the height they meet at will be the same. Therefore in the time it takes a to travel S m, b must have traveled 50+s m. This gives you two equations which you can then easily solve simultaneously. See if you can do it from there
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    (Original post by samb1234)
    Well the height they meet at will be the same. Therefore in the time it takes a to travel S m, b must have traveled 50+s m. This gives you two equations which you can then easily solve simultaneously. See if you can do it from there
    Yeah i found what T was that's 2.78.
    im having troubles with finding what h is these equations confuse me the most.
    If i plug in T to h=(2 x T) - (0.5 x -g x T^2)
    , i get a negative number for h which is wrong.
 
 
 
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