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Velocity Time Graph For Projectile!

  1. Feb 10, 2009 #1

    Hopefully I'm posting in the right section of the forum. I want to find out how the velocity-time graph (or any other graph for that matter) will look like for a projectile.

    Because I want to understand the theory behind projectiles properly and I can't seem to find any answers online or in my books.

    Will it be something similar to a cosine (since there is an initial velocity) but the shape will shift slightly cos of the initial angle to horizontal??

    Once I worked out that graph, then we'd break down the velocity into vertical and horizontal components and then work out all the unknowns.

    Then because we break down into components the graphs will be linear or something?? hence making it easier for us to work with and manipulate.

    Any feedback would be appreciated.
  2. jcsd
  3. Feb 10, 2009 #2
    Hello, I hope this will help.

    A velocity-time graph for a projectile fired at a 45 degree angle upwards we'll say, neglecting air resistance, would appear as a part of a parabola. As time continues, the projectile's velocity is decelerated by gravity. Eventually it becomes 0, and advances into negative figures until it hits the ground. It should be noted that a v-t graph is the derivative of a position time graph, and that an acceleration-time graph can be derived from the velocity-time graph.

    This is my elementary understanding of projectile motion, perhaps someone else could explain it better.
  4. Feb 10, 2009 #3


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    This is not correct. The graph you're describing is for the position-time graph of a projectile. The velocity-time graph would look line a straight line which intercepts the y-axis at a value v0 and has a constant slope of -9.8m/s^2. This is because in a projectile problem it is assumed that a=-9.8m/s^2, g, a constant. So when you integrate the acceleration function:
    [tex]\int\frac{dv}{dt}dt=\int -g dt [/tex]

    You get:

    [tex]v(t)=v_{0}-gt[/tex], which is, as you can see, a function with the characteristics I mentioned. Now what oscar wilde said is what is derived if you integrate once more:

    [tex]\int \frac{dy}{dt}dt=\int v_{0}-gt dt[/tex]

    Which is, as you can see, a parabola.

    If you haven't taken calculus, then this probably doesn't make any sense to you. But you can understand that if acceleration is the slope of velocity, and acceleration is constant, then the slope of the velocity-time graph must be constant (a straight line), which alone can lead you to deriving the velocity-time equation.

  5. Feb 10, 2009 #4

    Are you sure??? I know that acceleration due to gravity is constant..

    however the direction of the particle is constantly changing so is the graph really a straight line graph?
  6. Feb 10, 2009 #5


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    Yes, the graph of velocity-time is really a straight line! If you think about it, the projectile starts with an initial velocity v0. So this corresponds to some point on the positive v axis. Then, as you have said, since acceleration is constant and acceleration is the slope of the velocity-time graph, the slope must be -g. The point where the graph intersects the t-axis (v=0) is not where it hits the ground, but rather the time at which the projectile reaches the peak of its trajectory.

    Just keep in mind what the position-time, velocity-time, and acceleration-time graphs all represent and their respective shapes should make sense to you.
  7. Feb 10, 2009 #6
    Okay thanks.

    Now how would the graphs of the different components look like? ie horizontal and vertical components.. Would it look different to the one above?
  8. Feb 10, 2009 #7


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    Staff: Mentor

    If you have the projectile motion equations handy, you should be able to construct a graph yourself. Pick an initial velocity and launch angle, calculate x and y for various values of t, and plot them on a graph.
  9. Feb 10, 2009 #8
  10. Feb 17, 2009 #9
    Again are you sure about this? Because the direction is changing, so therefore the graph will not be straight (ie acceleration).

    However when resolved into x and y coordinates, then yes it would be straight.

    Anything wrong with this thinking??

    I remember that anything in orbit is in radial acceleration, because it's direction is changing.
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