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Homework Help: Projectile motion help

  1. Jul 18, 2009 #1
    Please tell me that in a projectile motion, how does the rate of change of speed equal to gcos[tex]\phi[/tex]
  2. jcsd
  3. Jul 19, 2009 #2

    Doc Al

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    What is Φ? In projectile motion, the acceleration is just g downward.
  4. Jul 19, 2009 #3
    I think you may be mixing some things up.

    For a projectile launched at an angle [tex]\phi[/tex] measured from the horizon, the axial projections of its velocities and accelerations are as follows, where [tex]V_0[/tex] is the initial velocity of the projectile):

    As for why this is the case, that's just breaking a vector down into its axial components.
  5. Jul 19, 2009 #4


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    From what you've given me, all I can say is what you're asking is how to derive:


    Now, I don't remember seeing any acceleration equations that are equivalent to this, and neither have I seen the velocity equation that corresponds to it:

    Taking the integral: [tex]v=gcos(\phi)t+c[/tex]

    So now I'm stuck, and mostly because as Doc Al has said,
    Yes, what is [tex]\phi[/tex]? How is it possibly being used as a variable in an acceleration equation?
  6. Jul 21, 2009 #5
    [tex]\phi[/tex] is the angle that the instantaneous speed makes with the horizontal at the instant we are measuring the change.
  7. Jul 21, 2009 #6
    If that's the case, I think the rate of change of speed is independent of [tex]\phi[/tex]
    Last edited: Jul 21, 2009
  8. Jul 21, 2009 #7
    That is correct. It is simply g.

    [tex]\frac{d\vec v}{dt}=\vec g[/tex]

    However, any breakdown of an acceleration vector into its axial projections, does take the angle [tex]\phi[/tex] into consideration.

    If we choose to adopt a coordinate system where the x axis coincides with the instantaneous velocity, then you can get the expression [tex]g\cos{\phi}[/tex]
    It is the instantaneous acceleration in the y direction (It is perpendicular to the instantaneous velocity and is pointing down).
    The component of the instantaneous acceleration in the x direction is [tex]g\sin{\phi}[/tex], opposing the direction of the motion.

    What's the framework of the question you need it for? I had to go out of my way to set up a situation where that expression had any meaning, so some clarification would be greatly appreciated.

    One other case of projectile motion where I've encountered that expression for the acceleration, is for the case of "projectile motion" restricted to a smooth inclined plane (Well, where I encountered it, we dealt with friction as well).
    Last edited: Jul 21, 2009
  9. Aug 14, 2009 #8

    George Jones

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    Isn't the rate of change of speed given by [itex]-g \sin \phi[/itex]?
  10. Aug 14, 2009 #9
    But the answer in the book is given to be gcos[tex]\phi[/tex]

    Could u please explain gj how u got -gsin[tex]\phi[/tex]???
  11. Aug 14, 2009 #10


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    I think George used the laws of projectile motion to get:


    where vx(t) and vy(t) are the horizontal and vertical velocities as a function of time and v0 is the initial speed. He then wrote speed as a function of time as speed(t)=sqrt(vx(t)^2+vy(t)^2). He then found d/dt of speed(t) and set t=0.
    Last edited: Aug 14, 2009
  12. Aug 14, 2009 #11
    Hi George Jones and supreabajaj

    Yes, it can be that if we take different axis reference as RoyalCat mentioned. Thx for correcting my mistake :)

    You can take a look at post #7 by RoyalCat ^^
  13. Aug 15, 2009 #12
    Please copy the question from the book, and not just the answer. None of us have a clear idea of what it is that you're asking.
  14. Aug 16, 2009 #13

    George Jones

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    As Dick suspected, I took the question at face value, and I calculated the rate of change of speed (magnitude of velocity), which is different from the magnitude of the rate of change of velocity. I did it two different ways.

    First way: differentiate with respect to time

    [tex]E = \frac{1}{2} m v^2 + mgy[/tex]

    keeping in mind that [itex]E[/itex] is constant.

    Second way: differentiate with respect to time

    [tex]v^2 = \vec{v} \cdot \vec{v}[/tex]

    keeping in mind that [itex]\vec{a} = - g \hat{y}[/itex].

    I think that the second way gives something like RoyalCat had in mind.
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