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Projectile Motion Using Coordinates

  1. Mar 16, 2008 #1
    1. A plane with velocity (350 m/s, 40 deg) dropped a rescue package at location (0, 300 m) and the package landed at location (range, 50m). Find the range and final velocity.



    2. d = vsin(40) x t + (1/2)x(a)x(t^2)
    dx = vcos(40) x t




    3. final velocity:
    first find the time in air:
    300 m= 350sin(40) x t + 1/2 (10 m/s^2)(t^2)
    t = 15 s
    final velocity = 350sin(40) + (10 m/s^2)(15s)

    300 m = t ((350sin(40) + (5 m/s^2)(t)) let t = 0
    t = 15 s

    final velocity = 350sin(40) + (10 m/s^2)(15s)
    = 375 m/s

    horizontal distance = 350cos(40) x (15 s)
    = 4021.73 m

    using coordinates (0, 300) and (?, 50) solve for ? using the distance formula:

    4021.73 m = sq root( (300-50)^2 + (0 - ?)^2))
    (4021.73 m)^2 = (300-50)^2 +(0-?)^2
    (1.6 x 10^7m) - (300 - 50)^2 = (-?)^2
    sq root (1.59375 x 10^7) = ?
    ? = 3392.18

    ...basically I would like to know if my math was correct and if I used the correct coordinates to solve for "?". Thank you for your help! :D
















     
    Last edited: Mar 16, 2008
  2. jcsd
  3. Mar 17, 2008 #2
    Looks Good.
     
  4. Mar 17, 2008 #3
    You really think so? The coordinates I used to solve for "range" were correct, right? By the way, thank you very much for taking the time to look at this problem. I really appreciate it :D
     
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