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Differentiate Velocity and pythagaros problem.

  1. Jul 17, 2015 #1
    1. The problem statement, all variables and given/known data

    A ball is dropped from a height of 100 ft, at which time its shadow is 500 ft from the ball. How fast is the shadow moving when the ball hits the ground? The ball falls with velocity 32 ft/sec, and the shadow is cast by the sun.


    If i draw this in a cartesian cordinate cordiante system: I place the height 100ft up the y-axis, the shadow is some place on the x axis, and the hypotenus or line connecting y and x is 500ft.





    2. Relevant equations

    How do i find the speed at which the shadow is moving when the ball hits the ground?

    3. The attempt at a solution

    I do know a couple of things:

    The distances y and x are related by phytagaros x^2 +y^2 =500^2


    If i isolate y at time t=0 i get
    y=sqrt(500^2-100^2)


    The speed of the ball must be the rate of change of distance x with respect to time t
    dx/dt =32 ft/s

    The speed of the shadow must be the rate of change of distance y with respect to time t
    dy/dt = unknown

    dy/dt must depend on dx/dt,

    dy/dt=dy/dx*dx/dt

    But here is my problem: When x=0 (when the ball hit's the ground)
    Then I have no idea what to differentiate? Also I can't differentiate dy7dt=d(sqrt(500^2-100^2))/dt

    A hint would be appreciated :)
     
  2. jcsd
  3. Jul 17, 2015 #2
    There is a related rates problem hidden in here.
     
  4. Jul 17, 2015 #3

    RUber

    User Avatar
    Homework Helper

    Be careful here. You really have ##x_0^2 + y_0^2 = 500^2 ##
    You can use this information to find the angle of the sun and turn y into a trigonometric relation to x at any height in the path.
     
  5. Jul 18, 2015 #4
    I do find this very difficult for some reason:

    First of all, just to argue why my first idea/logic must be wrong.
    If implicitely differenciate x0^2 +y0^2 =500^2 i could get dy/dt but this would be the speed of the shadow at t=t0, which is zero and i don't want that.

    2xdx/dt +2ydy/dt = 0
    dy/dt = 2x(dx/dt)/2y We know that at x=0 so the whole equation becomes is zero 2*0(dx/dt)/2*y = 0 right?

    So what is the hidden relationship?
    I know that sin^2 +cos^2 =1^2 and i know that y=r*sin and x=r*cos

    So i can find sin of omega by
    sin^(-1)(y/r)= 11.537 degree.

    How is y and x related in a right triangle`?
    I assume tan =y/x but how does this help me find dy/dt?

    I'm still not really seeing the picture.
     
  6. Jul 18, 2015 #5
    Remember, you are dropping the ball, so dy/dt is just the speed of the falling ball.

    dx/dt can be related to dy/dt by the trig function. The angle if the sun in the sky is (very nearly) constant while the ball is falling, so the ratio of dx/dt (horizontal shadow speed) to dy/dt (falling ball speed) is very nearly constant as the ball is falling,
     
  7. Jul 18, 2015 #6

    ehild

    User Avatar
    Homework Helper
    Gold Member

    The picture shows the falling ball and its shadows at different times. During the falling, the rays of the sun hitting the ball are parallel. What is the relation between y, hight of the ball, and x, horizontal coordinate of its shadow?
    shadows.JPG
     
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