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Projectile Motion glider

  1. Dec 2, 2009 #1
    1. The problem statement, all variables and given/known data

    A glider is tugged by an airplane at 81 m/s when it is released. If the original speed was along the horizontal and the glider is now under a constant acceleration of 2.4 m/s2 at 1.1° below the horizontal due to air drag, how long will it take to reach the ground 5.7 km below?

    a. 250,000 s
    b. 500s
    c. 4.8 s
    d. 2.2s

    2. Relevant equations

    v0x = v0*cos(theta)
    v0y = v0*sin(theta)
    vy = v0y + at
    x = x0 + v0x*t
    y = y0 + v0y*t + 1/2*at^2
    vy^2 = v0^2 +2a(delta y)

    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution

    No idea on how to start.
     
  2. jcsd
  3. Dec 2, 2009 #2
    Notice that the horizontal components of acceleration and velocity do not affect the time it takes to travel vertically - the time to reach the ground.

    y = y0 + v0y*t + 1/2*at^2

    This equation should work. You know y, y0, v0, and a. Solve for t.
     
  4. Dec 2, 2009 #3
    How do I find the initial acceleration in the y direction?
     
  5. Dec 2, 2009 #4
    The force of air drag ([tex]F_{drag}[/tex]) on the glider is in the opposite direction of the velocity of the glider. Since this drag is 1.1 degree below the horizontal, the [tex]F_{drag}[/tex] will have a horizontal and vertical component (i.e. will pull the glider backward and downward). So to find the acceleration of the glider in the y direction, we find the resultant force acting on the glider, which is the sum of the y components of the forces acting on it (e.g. [tex]F_{grav}[/tex] and [tex]F_{drag}[/tex]).

    [tex]F_{drag}[/tex] = 2.4 m/s2 at 1.1° below the horizontal
    [tex]y_{i}[/tex] = 5.7 km = 5700 m
    [tex]v_{i}[/tex] = 81 m/s

    [tex]\sum[/tex]F = ma
    [tex]\sum[/tex][tex]F_{y}[/tex] = [tex]F_{grav}[/tex] + [tex]F_{drag}[/tex] = m[tex]a_{y}[/tex]
    [tex]\sum[/tex][tex]F_{y}[/tex] = -mg - [tex]F_{drag}[/tex]sin(1.1) = m[tex]a_{y}[/tex]
    [tex]a_{y}[/tex] = -(g + ([tex]F_{drag}[/tex]sin(1.1))/m)

    From the y component of acceleration, you can derive the y equation as xcvxcvvc said, set that equation equal to zero since y = 0 at ground, and solve for t.
     
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