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    A particle is projected from a point c with horizontal and upward velocity 50ms and 30ms respectively find after how long particle is at same level as c
    How to get answer which is 6.12s
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    (Original post by Hasham123)
    A particle is projected from a point c with horizontal and upward velocity 50ms and 30ms respectively find afterglow long particle is at same level as c
    How to get answer which is 6.12s
    Can we have your working please?
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    (Original post by DreamlinerFinder)
    Can we have your working please?
    Sure
    Sv= UV x T + AT^2
    0=50T -4.9T^2
    Then put that into quadratic formula and got 10.204 which is wrong
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    (Original post by Hasham123)
    Sure
    Sv= UV x T + AT^2
    0=50T -4.9T^2
    Then put that into quadratic formula and got 10.204 which is wrong
    I think it was the quadratic formula step, but I can't remember how to do this, so I'll leave this for someone who knows it better

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    Use v = u +at for the vertical. For the time the ball is in the air is dependent only on the vertical not the horizontal.
    Arrange the formula to get t = (v - u)/a
    t = (30)/9.81
    t=3.058 This is the time taken for the particle to reach its max height so you will have to multiply by 2.
    3.058 x 2 = 6.12
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    s = ut + 1/2at^2

    0 = 30t - 4.9t^2

    0 = t(30 - 4.9t)

    t = 0 (not the answer you're looking for)

    30 - 4.9t = 0
    so t = 6.12s

    Your mistake is using 50ms instead of 30ms. That 50 is just there to mislead you.
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    Thank you from this I got -4.9t^2 + 30t =0 then put that in quadratic formula and got 6.12
 
 
 
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