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Finding the intersect of two functions

  1. Jul 28, 2014 #1
    I have the following equation which describes a water jets trajectory.

    Vx=Cos(50) * 6.5m/s
    Vy=Sin(50) * 6.5m/s

    x=Vx*t
    y=Vy*t-0.5*g*t^2

    where t=time

    Now, this plots a curve very well starting from zero.

    I want a line, starting from 0 also, that intersects the curve.
    Now, I have used the equation y=tan(35)*x where x is the distance across the bottom of the graph (x axis) on the adjacent part in the trig.

    So, to find the intersect, i need to get the curve equation in terms of x and y instead of y and t, so i have re-arranged; t=x/Vx and subbed in to give;

    y=Vy*(x/Vx)-0.5*9.81*(x/Vx)^2

    If I equal them to one another; tan(50) *x = Vy*(x/Vx)-0.5*9.81*(x/Vx)^2
    and re-arrange for x, i get;

    (tan(35)*Vx^2)/(-0.5*g)) - ((-Vx*Vy/(0.5*g)) = x = 0

    Now, there is an intersect at 0, but not the one i want.

    Attached is a graph of both the line and curve, with 50 degrees angle on the jet, and 35 degrees on the line.

    Please help, ive been stuck on this alllll day, and alot of yesterday too!
     

    Attached Files:

  2. jcsd
  3. Jul 28, 2014 #2

    HallsofIvy

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    That's a quadratic equation. It has two solution of which x= 0 is one. What is the other?
     
  4. Jul 29, 2014 #3
    Its not a quadratic equation because it cant be put into the form of ax^2+bx+c=0 since we have an x on both b and c, so what i have is ax^2+bx+cx=0.
     
  5. Jul 29, 2014 #4

    HallsofIvy

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    Which is equal to ax^2+ (b+ c)x= 0, a quadratic equation. As I said, x= 0 is one solution, what is the other.
     
  6. Jul 29, 2014 #5
    Okay, so where we have a =(0.5*9.81)/(Vx^2) = 0.2929
    b =Vy/Vx = 1.19175
    c=TAN(35) in degrees

    Plugging it all in and checking with calculator and excel, i get x=-0.7119 and x=-3.3568

    Neither of which are the intersection.
    Also, it is not referring to t or time, because the time range is only from 0-0.5
     
  7. Jul 29, 2014 #6
    In the previous comment, i assumed ax^2+bx+cx=0 since thats the form of my equation.
    If i calculate, by transforming my equation to ax^2+bx+c=0 by adding both a and b together (because they both have one x) and making it b, and then making c = 0, i get x=0 (promising) but then x=-6.459 which is incorrect.
    Driving me insane. I hate not being able to solve a problem.
    I will ask my tutor, but shes not in till next week. If anyone has any ideas then please comment.
     
  8. Jul 29, 2014 #7

    Ray Vickson

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    Check the sign of 'a'.
     
  9. Jul 30, 2014 #8
    Thanks, i tried subtracting c from b and i got the correct answer! It wasn't the sign of a, i had just simply got the signs wrong, but you pushed me in the right direction.
    Seems very simple now!
     
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