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Basic Projectile Motion Problem

  1. Aug 31, 2008 #1
    1. The problem statement, all variables and given/known data

    A physics student on PLanet Exidor throws a ball, and it follows hte parabolic trajectgory shown in the figure. The ball's position is shown at 1.00 intervals until t=3.00. At t=1.00, the ball's velocity is v=(2.00i + 2.00j)[m/s].

    [​IMG]

    a. Determine the ball's velocity at t=0, 2.00, and 3.00.
    b. What is the value of g on the Planet Exidor?
    c. What was the ball's launch angle?

    2. Relevant equations

    [tex]
    v = v_0 + a t
    [tex]

    [tex]
    x = x_0 + v_0 t + (1/2) a t^2
    [tex]
    [tex]
    v^2 = v_0^2 + 2 a \Delta x
    [tex]



    3. The attempt at a solution

    I was able to find at 1.00 that the ball's acceleration in the x direction = 2.00[m/s^2]. I'm not really sure how to find the velocity at t=0 since I think it can't be zero. I'm pretty if I can figure out the max height I can figure out what g is. But I can't find max height.

    Thanks.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 31, 2008 #2

    Kurdt

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    With all projectile motion problems you need to consider both components of motion. There should be no acceleration in the x direction. At the top of the parabola what do you expect the y component of velocity to be? This will help you with what you know about t = 1 to set up some simultaneous equations and solve the problem.
     
  4. Aug 31, 2008 #3
    It will really help to rewrite

    [tex]
    v = v_0 + at = v_0 - gt
    [/tex]

    where i've defined g downwards as positive as two equations then...

    [tex]
    v_x = ... \quad \& \quad v_y = v_y(t)=...
    [/tex]
     
  5. Aug 31, 2008 #4
    Thanks for the help. It was stupid of me to think of ax instead of just thinking g the whole time. This clarified things a bunch, thanks!
     
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