1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Relative Velocity and Projectile Motion

  1. Sep 19, 2008 #1
    1. The problem statement, all variables and given/known data
    A low-flying helicopter flying a constant 215 km/h horizontally wants to drop secret documents to his contact's open car which is traveling 155 km/h on a level highway 78.0 m below. At what angle (to the horizontal) should the car be in his sights when the packet is released?

    [​IMG]

    2. Relevant equations
    Kinematic Equations for Constant Acceleration in Two Dimensions
    X-component (horizontal):
    vx = vx0 + axt
    x = x0 + vx0t + .5axt2
    vx2 = vx02 + 2ax(x-x0)

    Y-Component (vertical):
    vy = vy0 + ayt
    y = y0 + vy0t + .5ayt2
    vy2 = vy02 + 2ay(y-y0)

    3. The attempt at a solution
    Knowns:
    Vhelicopter = 59.7 m/s
    Vcar = 43.1 m/s
    y = -78 m
    y0 = 0
    x0 = 0
    g = -9.8 m/s2

    vy2 = ?
    vy2 = vy02 + 2ay(y-y0)
    vy2 = 0 + 2 (-9.8 m/s2)(-78 m)
    vy = 39.1 m/s

    vx = Vhelicopter + Vcar = 59.7 m/s + 43.1 m/s = 102.8 m/s ? <-- I'm not sure I did this right. Probably didn't.

    tan Θ = vy/vx = 39.1/102.8
    Θ = 20 degrees

    But the back of the book says that the answer is 49.6 degrees. I'm probably wrong and somewhere in my work I screwed up somewhere. My calculator is in degree mode.
     
  2. jcsd
  3. Sep 19, 2008 #2

    Kurdt

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Ok try finding the x distance the parcel will travel and take the inverse tan of the ratio of the two sides (i.e. x - distance and y - distance).
     
  4. Sep 19, 2008 #3
    All right. I'm not sure if this is correct or not, but I found time.

    y = y0 + vy0t + .5ayt2
    -78 m = 0 + 0 + .5 (9.8 m/s2)t2
    t = 3.99 s

    Then after I found time. I took your advice.

    x = x0 + vx0t + .5axt2
    x = 0 + (59.7 m/s - 43.1 m/s)(3.99 s) + 0
    x = 66.2 m

    tan Θ = y/x = 78 m/66.2 m
    Θ = 49.6 degrees

    I'm still a bit confused about relative velocity though. Can someone explain how relative velocity works?
     
  5. Sep 19, 2008 #4

    Kurdt

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

  6. Sep 19, 2008 #5
    Thank you so much for your help!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Relative Velocity and Projectile Motion
Loading...