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How Tall Is The Building? - Projectile Motion

  1. Oct 4, 2009 #1
    1. A 0.43 kg rock is projected from the edge of the top of a building with an initial velocity of 9.2 m/s at an angle 45◦ above the horizontal. Due to gravity, the rock strikes the ground at a horizontal distance of 14.9 m from the base of the building.

    How tall is the building? Assume the ground is level and that the side of the building is vertical. The acceleration of gravity is 9.8 m/s2 .




    2. Relevant equations



    3. The attempt at a solution
    I've broken the problem down into halves. I've solved the top half ( the part of the motion equal in height to the roof ) but don't know how to get the second half ( the part of the motion after the rock is past the point level with the roof. ) Have I even gone about this right?

    First half:
    Vxi=9.2cos45=6.51m/s Vyi=9.2sin45=6.51m/s

    delta t=(Vfy-Viy)/g =1.33 s

    delta d=Vix(delta t) =8.65 m

    Second half: (Not sure if right)
    Vix=6.51cos45=4.6m/s Viy=6.51sin45=4.6m/s

    Now I am stuck!
     
  2. jcsd
  3. Oct 4, 2009 #2
    You can fit all the information for the Y axis in to one equation:

    d=Vot + 1/2at2

    So d is your vertical displacement, Vo your initial velocity, etc. Make sure you use proper signs, otherwise this will fail horribly.

    Graphing is never accurate, so let's go about this algebraically.

    The things you don't know are t, and d.

    So if you find t you can find d.

    You know the horizontal displacement, and the horizontal velocity, so by using:

    Dx=Vxt

    you can solve for t, then put it into the first equation.

    Your resulting displacement should be negative. The absolute value of this is the height of the building.

    If you did move d over and make the whole thing a quadratic equation, the graph would exactly model this setup.

    Give it a try.
     
  4. Oct 4, 2009 #3
    [tex]\vec{a}(t)=\left(\begin{array}{c} 0 \\-g \end{array}\right)[/tex]

    by integrating this expression we get:

    [tex]\vec{v}(t)=\left(\begin{array}{c} v_0\cos\alpha \\-gt +v_0\sin\alpha\end{array}\right)[/tex]

    and integrating a second time gives

    [tex]\left(\begin{array}{c}x(t) \\y(t)\end{array}\right)=\left(\begin{array}{c} v_0t\cos\alpha \\- \frac 1 2 gt^2 +v_0t\sin\alpha +h\end{array}\right)[/tex]

    hence

    [tex]y=-\frac{g}{2v_0^2 \cos^2 \alpha} x^2 + x\tan \alpha +h[/tex]

    and finally

    [tex]y=0 \Rightarrow h= \frac{g}{2v_0^2 \cos^2 \alpha} x^2 - x\tan \alpha}[/tex]

    [tex]v_0=9.2 m.s^{-1},\alpha=45^{\circ}{, x=14.9m[/tex]
     
  5. Oct 4, 2009 #4
    Okay, so the values I put in for Dx and Vx are 8.65 m and 4.6 or 6.51?
     
  6. Oct 4, 2009 #5
    What is the X component of 9.2 m/s at 45 degrees?

    9.2cos45 = 6.51

    Where did you pull 4.6 from?
     
  7. Oct 4, 2009 #6
    Ha, the second part of my first attempt at this...I didn't know if I should even have done that. Sorry.
     
  8. Oct 4, 2009 #7
    No need to apologize.

    Just follow the steps I outlined and you'll get it.

    If you want me to explain more feel free to ask questions.
     
  9. Oct 4, 2009 #8
    So, I get 1.33 s for t. Now, putting that value for t into the first equation, I get 3.57 for d. I thought it was supposed to be negative?
     
  10. Oct 4, 2009 #9
    I got 2.218 seconds.

    Did you do your algebra correctly?

    The displacement I got was 38.9 meters.

    I should have been more specific, if you're putting in a distance then it should be negative, but since you're finding it it'll come out positive.
     
  11. Oct 4, 2009 #10
    Okay, maybe I'm still not understanding. I did this:
    Dx=Vxt
    8.65m=6.51m/s(t)
    8.65m/6.51m/s=t
    1.33=t

    Was I supposed to?
     
  12. Oct 4, 2009 #11
    A 0.43 kg rock is projected from the edge of the top of a building with an initial velocity of 9.2 m/s at an angle 45o above the horizontal. Due to gravity, the rock strikes the ground at a horizontal distance of 14.9 m from the base of the building.

    Dx should be 14.9

    Vx should be 9.2cos(45)

    Dx=Vxt

    So to find time

    t=Dx/Vx
     
  13. Oct 4, 2009 #12
    Yes, now I get 2.28 s
     
  14. Oct 4, 2009 #13
    So shove it into the other equation and see what you get.
     
  15. Oct 4, 2009 #14
    I get d= -4.69
     
  16. Oct 4, 2009 #15
    Check your calculations. I got -10.63988816.
     
  17. Oct 4, 2009 #16
    I still can't get that as an answer.
     
  18. Oct 4, 2009 #17
    Does your equation look like this?

    d = 9.2cos(45)t + .5(-9.8)t2

    d = 6.50538(2.28) + -4.9(2.28)2

    d = 14.83227 + (-25.47216)
     
  19. Oct 4, 2009 #18
    Oh, I just used the initial velocity, not the x component of it.
     
  20. Oct 4, 2009 #19
    AGH! NO! WRONG!

    You use the Y component, the only equation for X velocity in a projectile motion scenario is Dx = Vxt

    That is solving for parts of the Y component. In this case the X and Y components just happen to be the same because of the angle.

    My bad, it should've said sin(45).

    d = 9.2sin(45)t + .5(-9.8)t2

    d = 6.50538(2.28) + -4.9(2.28)2

    d = 14.83227 + (-25.47216)
     
  21. Oct 4, 2009 #20
    Oh, okay. So yes, now I get the same answer as you. My apologies, I am very slow at this stuff.
     
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