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    A ball is projected from a point on the horizontal ground. The speed of projection is 30ms-1, greatest height reached is 20m. Assuming no air resistance, find angle of projection above horizontal.
    - And it also goes on to say, find the speed of the ball as it passes through highest point. ) - how can it past through the height-est point

    i tried using V=u+at, i ended up with t=30sinX/9.8 then i tried to find t for the horizontal, which i got t=x/30cosx, equating both (T)
    and solving by simultaneous equations doesn't help.
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    at the highest point the vertical component of velocity is zero... so u2 + 2as = 0

    here u is the initial vertical component of velocity = Vsinθ = 30sinθ

    a = -9.8 ms-2

    s = height = 20 m.
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    (Original post by the bear)
    at the highest point the vertical component of velocity is zero... so u2 + 2as = 0

    here u is the initial vertical component of velocity = Vsinθ = 30sinθ

    a = -9.8 ms-2

    s = height = 20 m.
    yh and then i used the equation v=u+at, t=\dfrac{30sin \theta}{9.8} for vertical

    i guess i should substitute this into s=ut+ 1/2 at^2
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    (Original post by joyoustele)
    yh and then i used the equation v=u+at, t=\dfrac{30sin \theta}{9.8} for vertical

    i guess i should substitute this into s=ut+ 1/2 at^2
    if you know t then you can use those... my version does not need t.
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    (Original post by the bear)
    at the highest point the vertical component of velocity is zero... so u2 + 2as = 0

    here u is the initial vertical component of velocity = Vsinθ = 30sinθ

    a = -9.8 ms-2

    s = height = 20 m.
    Oh i see, that is a more simple way of doing it. Thank you
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    (Original post by joyoustele)
    Oh i see, that is a more simple way of doing it. Thank you
    your method works fine, and quite often you need to know t at the top anyway.
 
 
 
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