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Homework Help: Kinematics in two Dimension- projectile motion

  1. Jun 3, 2007 #1
    I am really having trouble solving this problem. Can someone help ASAP??

    An airplane with a speed of 87.9 m/s is climbing upward at an angle of 59.8 ° with respect to the horizontal. When the plane's altitude is 652 m, the pilot releases a package. (a) Calculate the distance along the ground, measured from a point directly beneath the point of release, to where the package hits the earth. (b) Relative to the ground, determine the angle of the velocity vector of the package just before impact.
  2. jcsd
  3. Jun 3, 2007 #2
    treat this as an elastic explosion
  4. Jun 3, 2007 #3

    Doc Al

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    Staff: Mentor

    Start by finding the horizontal and vertical components of the initial velocity of the package.
  5. Jun 3, 2007 #4
    This is what i did:
    I first drew a right triangle from the start to the altitude. I got x = 379 m.

    I then calculated the x and y components of velocity:
    V_ox=87.9cos(59.8)=44.2 m/s
    v_oy=87.9sin(59.8)=76 m/s

    I know that a_x= 0 m/s^2
    a_y = -9.8 m/s^2
    v_ymax= 0 m/s
    y=652 m
    x_max ??

    I then calculated time by using v_y = v_oy + a_y*t
    t= -v_oy / a_y = 7.8 s (2) because this was only half the length of x_max

    After I got t, I used the equation x_max=v_ox*t + 1/2a_x*t^2
    --------------------------------------------------^ this part equals to 0
    so I got x_max = 1204.3 m - 379 = 825.3 m
    I am not sure if I am going the right path.

    Next to calculate the angle, I set v_ox = 44.2 m/s
    a_x = 0 m/s^2
    y=652 m
    v_oy = 0 m/s
    a_y = -9.8 m/s^2

    I used the equation v_y^2 = v_oy^2+2a_y*y
    v=45.5 m/s

    pheta = cos^-1(v_x/sqrtv_x^2+v_y^2)
    = 12.0 degrees

    Could someone tell me if I followed the right procedures and if my answers are right or wrong?
  6. Jun 3, 2007 #5
    so i really have no idea what you did but i'm going to walk you through how to get the right answer.

    1. you got the components and initial height correct
    2. now you have your components in the x direction and the y direction and you need to find out how long it takes for the negative acceleration sap all the velocity from the package, you got this part right and the fact that the package travels for twice that distance.
    3. i have no idea how you figured that the distance it travels from release and back down to the ground.

    the way to do it is take the first [tex] 2 \times 7.8s [/tex] and multiply that by it's [tex]V_x[/tex] and i get 685m. now it's at position (379+685, 652) and heading downwards and you have to find out how long it takes to get from there to y=0 or y=-652.

    this is done using the position function

    [tex] -652 = V_yt - \frac{1}{2}9.81t^2 = -76t -4.905t^2 [/tex]
    important thing to know is that it comes down with the same [tex]V_y[/tex] with which it goes up but negative, so -76m/s. another thing to notice is that i've made the 652m altitude my origin, which is why everything is negative. now this is a quadratic equation solved either with the quadratic equation or by graphing, i graphed it and got t= 6.144s for it to drop the final 652m. multiply this by its [tex]V_x[/tex] and you got 271.5m. total distance from lift off to drop is 379+685+271.5 = 1335m

    now for the vector you need to figure out the component velocities again and do the trig. i'll let you try that
    Last edited: Jun 3, 2007
  7. Jun 3, 2007 #6


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    Homework Helper

    The question says
    The 379m is not part of this.
  8. Jun 3, 2007 #7
    that is why i made it very transparent in my answer that i did include it in my final number
  9. Jun 3, 2007 #8


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    Homework Helper

    It shouldn't be included in the final number. The distance from the point of release to when it hits the ground is 685 + 271.5 = 956.5m. The 379m is the distance from liftoff up to 652m, which is not part of the distance we are asked to find, so it is irrelevant.

    By the way, I think your signs for v and a are mixed up in your quadratic equation.
  10. Jun 3, 2007 #9
    dude seriously , it's not that big of a deal, i chose to include the initial distance so he would know where in space the projectile was.

    1 sign was wrong before all the numbers were plugged in, but the pertinent signs were all correct.
  11. Jun 3, 2007 #10
    Thanks for all your help and input. I found the answer online on cramster.com and they showed how to do the problem step by step.
  12. Jun 3, 2007 #11
    your exact problem?
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