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Projectiles anyone?

  1. Nov 14, 2005 #1
    A stone is thrown vertically upwards with an initial velocity of 19.6m/s. What is it's velocity and position after 1s, 2s, 3s?
    I'm getting confused between my positives and negatives. What answers did you get?
  2. jcsd
  3. Nov 14, 2005 #2
    First, write what you know:

    Initial Velocity = 19.6m/s
    acceleration = -9.81
    Time = 1,2,3

    So, from the formula d=Vi(t)+.5a(t)^2, we get:

    di = 19.6(1) + (.5)(9.81)(1)^2
    dii = 19.6(2) + (.5)(9.81)(2)^2
    diii = 19.6(3) + (.5)(9.81)(3)^2

    di = 24.505m
    dii = 58.82m
    diii = 102.945m

    And now, Vf^2 = Vi^2 + 2ad:

    Vfi = Root[(19.6^2) + 2(-9.81)(24.505)]
    Vfii = Root[(19.6^2) + 2(-9.81)(24.505)]
    Vfiii = Root[(19.6^2) + 2(-9.81)(24.505)]

    I don't have a calculator on hand, but these should work out to be the proper velocities, unless I messed up somewhere. :wink:
    Last edited: Nov 14, 2005
  4. Nov 14, 2005 #3


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    Your answers are incorrect.

    Initial velocity was upwards - gravity accelerates objects downward. Positive is upwards and negative is downwards. (In fact, you knew that. Unfortunately you didn't enter it into your equation that way.)

    The velocity is just the derivative of the position equation:

    [tex]v_f = v_i + at[/tex]

    *acceleration is still downwards.
  5. Nov 14, 2005 #4
    Yeah, I see what I did. Arithmetic always seems to kill me.
  6. Nov 14, 2005 #5
    Bob G can you help me - getting so confused? What answers did you get because mine don't make sense
  7. Nov 14, 2005 #6
    Fabbo, sorry to bumb in like this but what have you done up till now. Please share us your results and how you got to them. Have you already consulted our marvellous tutorials ?

    This is the best way to learn these important introductory concepts of Newtonian Physics.

  8. Nov 14, 2005 #7
    ok here goes.......

    s = ut + 1/2 at^2

    so after 1 sec

    s = 19.6 x 1 + 1/2 x 9.81 x 1^2
    s = 14.7m

    v = u + at

    so after 1 sec

    v = 19.6 + -9.81 x 2
    v = 9.79 m/s

    I did the rest like this and got answers for 2 seconds as - 0.02m/s and 19.58m and for 3 seconds - 9.83 m/s and 14.655m

    I don't think they're right and I don't understand when the stone is going up and falling .............
  9. Nov 14, 2005 #8
    Now we are getting started.
    First of all, like you can see in the tutorial that i referred to, the above formula consists out of VECTORS. The clue is to express the components of these vectors with respect to the x and y-axis. For example, in this case, there is one acceleration : GRAVITY.
    Acceleration is a vector and gravity is directed in the opposite direction (downward) of the y-axis. So you have
    [tex]\vec{a_y} = -g \vec{e_y}[/tex]

    The initial velocity is directed upwards in the same direction as the y-axis , so you have :
    [tex]\vec{v_{initial}} = 19,6 \vec{e_y}[/tex]

    WATCH THE SIGNS and be sure when to use + and - (same <--> opposite direction of the y-axis)

    Now, try to write down the formula's for position and velocity in both the x and y direction.

    The x-component of both acceleration and velocity are 0. Do you realize this ?

    I strongly suggest you read the TUTORIAL. You will find more indept info on this along with some examples/exercises


    The general formula's for velocity (vector v) and position (vector r) are :

    [tex]\vec{v} = \vec{v_0} + \vec{a}t[/tex]
    [tex]\vec{r} = \vec{r_0} + \vec{v_0}t+ \vec{a} \frac{t^2}{2}[/tex]
    Last edited: Nov 14, 2005
  10. Nov 14, 2005 #9


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    Yes, they are right (you're using the gravitational acceleration constant in your calculator, aren't you).

    To find where the dividing line between going up and going down is, the second equation (v=u + at) should equal zero at that the precise instant between going up and coming down. If you know the initial velocity and the acceleration, you can find rearrange to find the time.
  11. Nov 14, 2005 #10

    No they are NOT correct. Look at the equation for the position, more specifically, look at the acceleration !

  12. Nov 14, 2005 #11
    does that make the first velocity 9.79m/s?

    how does this fit into the position equation? i got 4.9m but was pretty confused....
  13. Nov 14, 2005 #12
    ok, how about this

    v = 19.6 -9.81t
    y = 19.6t -9.81t²/2

    Try these...


    i must say that you should be able to come up with these yourself.
  14. Nov 14, 2005 #13
    first velocity is indeed 9.79 m/s

    first position 14.695 m
  15. Nov 14, 2005 #14
    oh ok i understand now, thank you for your help - it really is appreciated. I know to an experienced physicist I should be capable of doing these. But unfortunately I'm not and that is what is so great about this forum. I'm not looking for people to tell me answers, I just want to further my understanding. I have already asked my school teacher and I did not find he was able to help me. Therefore thank you and your tutorials are brilliant.
  16. Nov 14, 2005 #15

    if you have any other questions on this, please do not hesitate. I just wanna stress (but i am sure many people have told you this) that it is very important that you know how to work with vectors/components, etc...

    Good Luck



    edit : now i am off too, to see LOST on tv
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