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Projectile Motion with displacement and time

  1. Apr 18, 2012 #1
    1. The problem statement, all variables and given/known data
    A boy kicks a ball into the air. It takes 4.3 seconds to land 41.1 meters from his position to the right. What is the launch velocity and launch angle of the ball? Ignore air resistance.

    Δx = 41.1 m
    t = 4.3 s
    a = -9.8 m/s2


    2. Relevant equations

    dx = vxt
    vfy = v0y + at

    3. The attempt at a solution

    dx = vxt
    dx / t = vx
    41.1 m / 4.3 s = vx = 9.56 m/s

    vfy = v0y + at
    0 = v0y + (-9.8 m/s2 * 4.3 s)
    0 = v0y - 42.1 m/s
    v0y = 42.1 m/s

    launch velocity = (9.56 m/s)2 + (42.1 m/s)2 = vi2
    vi = 43.2 m/s

    launch angle = vxi = vicosθ
    9.56 m/s = 43.2 m/s * cosθ
    θ = 77.2°

    Can anyone please look through my work to see if I have done it correctly? I'm getting conflicting answers from wolframalpha.

    Thanks!
     
    Last edited: Apr 18, 2012
  2. jcsd
  3. Apr 18, 2012 #2
    Looks all good to me.
     
  4. Apr 18, 2012 #3
    It seemed like that to me, yet when I input these values into http://www.wolframalpha.com/input/?i=projectile+motion, I get travel time that is twice the question's (8.6s) and a horizontal displacement that is also twice the distance (82.2m).

    I can't figure out why there's a difference.
     
  5. Apr 18, 2012 #4

    gneill

    User Avatar

    Staff: Mentor

    Okay, that looks good.
    Oops. The vertical velocity won't be zero when it lands --- it's going to impact the ground at some speed (and probably bounce). (When is the vertical velocity zero?)

    You can use this kinematic relationship if you choose the appropriate time when the velocity is zero, otherwise find another equation for the vertical vertical trajectory that includes the landing condition (Hint: height at landing is zero).
     
  6. Apr 18, 2012 #5
    Whoops, made a mistake in my calculations. You should probably try d = vt + 1/2at2 instead. As gneill just said, your Vfy is 0 at the TOP of it's trajectory, not the bottom.
     
  7. Apr 18, 2012 #6
  8. Apr 18, 2012 #7
    Ahhh, I see. I'm calculating AT time 4.3s!

    dy = 0 = vt + 1/2at^2

    Better?
     
  9. Apr 18, 2012 #8
    :approve: yesh
     
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