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Math Help - projectiles

  1. #1
    Member
    Joined
    Sep 2008
    Posts
    239

    projectiles

    A stone is projected in a direction which makes an angle of 45° above the horizontal. It strikes a small target whose horizontal and vertical distances from the point of projection are 120m and 41.6m respectively. The target is above the level of the point of projection.

    a) Find the speed of projection and show that the time taken for the stone to reach the target is 4s.

    b) Determine, correct to 2d.p., that the speed and direction of motion of the stone as it hits the target.

    Ok i have been taught very little on projectiles.

    I have these to play around with (column notation)

    when t=0

    a =
    0
    -g

    or
    (0i-gj)

    V=
    vcosα
    -gt+vsinα

    or
    ((vcosα)i + (-gt+vsinα)j)

    r=
    (vcosα)t
    (-gtē)/t + (vsinα)t

    or
    (((vcosα)t)i + ((-gtē)/t + (vsinα)t)j)


    not really sure how to start the question
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  2. #2
    Member u2_wa's Avatar
    Joined
    Nov 2008
    Posts
    119
    Hello djmccabie

    You must know that horizontal speed remains constant(if air resistance is ignored)
    Vertical velocity =v sin(45)
    Horizontal velocity =v cos(45)

    Horizontal distance =120=v cos(45).t(t=time period)

    Vertical distance =41.6= v sin(45).t-\frac{1}{2}(9.81)(t^2)
    ( s=ut+\frac{1}{2}at^2)

    Solve both equations to get v&t

    b) Vertical speed =v sin(45).t-9.81t=v sin(45).4-9.81(4)
    Horizontal speed =v cos(45)
    Resultant vel. at that point =\sqrt{(vel.vertical)^2+(vel.horizontal)^2}

    Direction of motion =arctan(\frac{vel.vertical}{vel.horizontal}) angle to the horizontal

    Hope this helps
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