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    A ball is thrown so that it goes as high as it goes forward. At what angle is it thrown?

    Here is my answer:

    a = Angle of projection
    V = Initial Speed of projection

    At this point where the ball reaches its max height, and it's vertical height is equal to its horizontal motion forward:

    x = Vcosat
    y = t(Vsina - 4.9t)

    Knowing that at this height, its velocity will be zero:

    0 = Vsina - 9.8t
    V = (9.8t)/sina

    At this particular point in the air:

    Vcosat = t(Vsina - 4.9t)

    Substitute the value for V into the above equation:

    [(9.8t)/sina]cosat = t([(9.8t)/sina]sina - 4.9t)
    (9.8t^2cosa)/sina = t(9.8t - 4.9t)
    (9.8t^2)/tana = 9.8t^2 - 4.9t^2 = 4.9t^2

    (9.8t^2)/tana = 4.9t^2
    tana = (9.8t^2)/4.9t^2
    tana = 2
    a = tan^-1 (2)
    x = 63.4 degrees (3.S.F)

    The angle the ball is thrown at is 63.4 degrees (3.S.F.)

    Is this correct? Appreciate responses
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    The ball travels in a parabolic curve. When it reaches the top of the curve, it is at its maximum height ,but has only travelled half its horizontal distance.

    Your line,

    At this particular point in the air:

    Vcosxt = t(Vsinx - 4.9t)


    should be,

    At this particular point in the air:

    Vcosxt =1/2*t(Vsinx - 4.9t)
    (i.e horizontal dist = half vertical dist)

    all the calculations looked ok otherwise.
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    (Original post by Fermat)
    The ball travels in a parabolic curve. When it reaches the top of the curve, it is at its maximum height ,but has only travelled half its horizontal distance.

    Your line,

    At this particular point in the air:

    Vcosxt = t(Vsinx - 4.9t)


    should be,

    At this particular point in the air:

    Vcosxt =1/2*t(Vsinx - 4.9t)
    (i.e horizontal dist = half vertical dist)

    all the calculations looked ok otherwise.
    Cheers
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    I got the angle =75.96 degrees.
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    (Original post by Fermat)
    I got the angle =75.96 degrees.
    yep.
 
 
 
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