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How I can get the Velocity from knowing the distance

  1. Feb 16, 2007 #1
    Please help!!

    I need to figure this out for my lab due today and I just can't get it, I appreciate any input.

    Using the folowing data I have to find the average Velocity in the x for the projectile and determine the Velocity.

    Angle of Gun Distance in meters
    0 degrees 5.795 meters
    20 degrees 6.10 meters
    27 degrees 7.32 meters
    30 degrees 7.625 meters
    37 degrees 7.778 meters
    40 degrees 8.083 meters
    47 degrees 7.869 meters
    50 degrees 7.717 meters


    I don't understand how I can get the Velocity from knowing the distance in the x and the angle but I think I am supposed to be using Vx = Vcos(theta)

    This is all the information I was given and I would be really really gratiful if someone could help me with figuring out how to get the velocity. Thank you.
     
  2. jcsd
  3. Feb 16, 2007 #2

    cristo

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    You need to find an equation which x as a function of v. The aim of this type of question is to use the vertical components to eliminate a variable from the horizontal equation for distance.

    So, what equations do you know? What do you know about the motion in the vertical direction; more specifically, what is the total displacement in the vertical direction? What is the equation for displacement in the x direction?
     
    Last edited: Feb 16, 2007
  4. Feb 16, 2007 #3
    well if I use Vf^2 = Vi^2 +2ad I could set Vf = zero because at the projectile's maximum height there is no velocity in the y compoent. Then after I have Vi in the y I could plug into Vy = Vsin(theta)

    would that work? Is the Vy in the latter equation the same as Vi in the y?

    displacement in the x direction is dx = Vix(t) but I don't know what time is.


    Thank you soo much.
     
  5. Feb 16, 2007 #4

    cristo

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    You could do, but you don't know y (max height), so that probably won't work

    This is correct for x. The way I was thinking of was this: You know the total displacement in the y direction (since the ball is shot off the ground, and the hits the ground at a certain point further on). So, you (should) know an equation for y in terms of t, a and vyi. You know a and y, and want to find vyi, so write this equation in a way so that you can eliminate t from x=vixt.
     
  6. Feb 16, 2007 #5
    oo dy= 1/2 a t^2 so then 0 = -4.9t^2 but then that gives me time is equal to zero. And I can't set dy=dx.

    Or if I use the whole equation d = Vi(t) + .5at^t and then rearrange it to
    -Vi(t) = -4.9t^2 divide both sides by -t and get Vi=4.9t.

    Then maybe set it equal to t and set dx=vix(t) to t so I can set the two equations equal to each other and get Vi/4.9=dx/Vix with dx known... but isn't that mixing the x and y components??

    This should be easy I am just in my first year of physics in 11th grade.....

    Either way thank you for helping me so much. You're the only person keeping me from failing right now.
     
  7. Feb 16, 2007 #6

    cristo

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    this isn't right, as your equation is not complete!

    This is more like it.
    You're correct that you cannot mix up x and y components (apart from t, which is of course the same for both components).

    So, you have viy=4.9t, and x=vixt. Now, can you express viy and vix in terms of V and θ (the velocity and the angle the velocity makes with the horizontal, respectively).

    If so, you can substitute these into the two equations, and then eliminate t.
     
  8. Feb 16, 2007 #7
    so set viy= Vsinθ and Viy =4.9t equal to each other to get Vsinθ=4.9t

    and then Vix=dxt and Vx=Vcosθ to get dx(t)=Vcosθ but how do I elimante t?

    I don't follow this last part....
     
  9. Feb 16, 2007 #8

    cristo

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    Your second equation is incorrect. It should read x=(Vcosθ)t. So, you then have Vsinθ=4.9t, which gives t=(Vsinθ)/4.9. Substitute this into x=(Vcosθ)t and you will obtain an equation for V which you can solve.
     
    Last edited: Feb 16, 2007
  10. Feb 16, 2007 #9
    dx=(Vcosθ)((Vsinθ)/4.9) whew that was a mouthful. Thank you so much.
     
  11. Feb 16, 2007 #10

    cristo

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    You're welcome.
     
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