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Math Help - determine force and height.

  1. #1
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    determine force and height.

    A projectile of mass 1.050 kg is shot straight up with an initial speed of 24.0 m/s.

    (a) How high would it go if there were no air friction?
    in meters

    (b) If the projectile rises to a maximum height of only 23.2 m, determine the magnitude of the average force due to air resistance.
    in Newtons
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  2. #2
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    Quote Originally Posted by rcmango View Post
    A projectile of mass 1.050 kg is shot straight up with an initial speed of 24.0 m/s.

    (a) How high would it go if there were no air friction?
    in meters

    (b) If the projectile rises to a maximum height of only 23.2 m, determine the magnitude of the average force due to air resistance.
    in Newtons
    (a)

    (v_{f})^2=(v_{i})^2+2ad

    0=24.0^2+2(-9.81)d

    0=576+(-19.62)d

    -576=-19.62d

    \frac{-576}{-19.62}=d

    29.36=d

    not so sure about b but i think you could factor the air resistance into acceleration because its average force so ill try it that way

    (v_{f})^2=(v_{i})^2+2ad

    0=576+2a(23.2)

     -576=46.4a

     -12.41=a

     -12.41-(-9.81)=-2.6

     f=ma

     f=(1.050)(-2.6)=-2.73n
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  3. #3
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    Quote Originally Posted by rcmango View Post
    A projectile of mass 1.050 kg is shot straight up with an initial speed of 24.0 m/s.

    (b) If the projectile rises to a maximum height of only 23.2 m, determine the magnitude of the average force due to air resistance.
    in Newtons
    Here's a more systematic approach to b) since you are doing energy problems.

    We know there is air resistance, so we know that there are non-conservative forces lurking about. Thus
    W_{nc} = \Delta E

    The force due to air resistance (F) is always in the opposite direction of the displacement, so
    W_{nc} = \vec{F} \cdot \vec{s} = Fs \cdot cos(180^o) = -Fs

    So
    -Fs = \frac{1}{2}mv^2 - \frac{1}{2}mv_0^2 + mgh - mgh_0

    Call upward positive and set the zero point for the gravitational potential energy at the point where the projectile is shot upward. (Typically I choose the 0 point for GPE at the lowest point in the diagram. That way I don't have to mess around with negative h values.) So h_0 = 0~m, and since the projectile is going to the max height, v = 0 m/s. Also note that s = h in this coordinate system. Thus
    -Fh = - \frac{1}{2}mv_0^2 + mgh

    F = \frac{mv_0^2}{2h} - mg

    You can plug the numbers in.

    -Dan
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  4. #4
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    Thankyou for the work here, I didn't realise that i double posted this problem.

    I appreciate the solutions.
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