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Projectile Motion Using Speed Graph

  1. Oct 2, 2008 #1
    1. The problem statement, all variables and given/known data
    A golf ball is struck at the ground level. The speed of the golf ball as a function of time is shown in figure where t = o at the instant the ball is struck. What is the initial velocity of the ball in unit vector notation? How far does the golf ball travel horizontally? What is the maximum height?

    2. Relevant equations
    a^2 + b^2 = c^2

    3. The attempt at a solution

    I would just use the range and height equations for parts 2 and 3 of the question. But I'm really stuck at part 1.

    From the graph, initial velocity is 20, but I'm not really sure what else to do. I have no theta, so I can't break it down into component vectors. From the solution key:

    Vx = 5 and Vy = 19.36

    How could you get Vx from this graph? Also, what is this graph showing? Speed never reaches zero. It would have to, though, wouldn't it? Also, it doesn't look linear...it curves a bit at 2. Does that imply that acceleration isn't constant?

    I think I just got it. Speed never reaches zero because Vx=5 when Vy=0. And since Vx final = Vx initial, Vx=5. Is my thinking correct? I'm still unsure about the curve in the graph though.
    Last edited: Oct 2, 2008
  2. jcsd
  3. Oct 2, 2008 #2


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    Its acceleration is decreasing, then is 0 at 2s and increases again after that.
    Try using some of Newton's equations of motion (e.g. s = ut + 1/2at2)
    I'm not sure if they will work though because acceleration isn't constant.

    What kind of physics coures are you doing and what have you been doing in class. That might give you a hint.
  4. Oct 2, 2008 #3
    I'm only taking Physics 1, non honors.

    I also thought acceleration wasn't constant because it looks like the graph is non linear. But in the solution key, the standard kinematic equations are used (range and height) so I'm a bit confused.
  5. Oct 2, 2008 #4


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    Well a=(v-u)/t
    a = (15 - 20)/0.5 = -10ms2
    a = (5 - 10)/1 = -5ms2
    So acceleration is changing. Try using Newton's equations and see what you get.
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