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Math Help - Speed of shadow.

  1. #1
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    Speed of shadow.

    Hi guys,
    I'm stuck on this problem.

    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.

    The only thing I have so far is that the derivative of the height of the ball is -32. I have no idea where to go from there.
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  2. #2
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    Re: Speed of shadow.

    calculate the time in which ball hits the ground, and from that information u can get the akceleration of the shadow rest is easy. sry dont have much time maybe later i repost.
    Last edited by Imo; July 5th 2012 at 11:41 AM.
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  3. #3
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    Re: Speed of shadow.

    Quote Originally Posted by BrettSGLane View Post
    A ball is dropped
    ... not projected, so initial velocity = 0.

    Quote Originally Posted by BrettSGLane View Post
    at which time its shadow is 500 ft from the ball.
    I suspect this was 500 ft from the point on the ground directly under the ball, since the question is about the velocity of the shadow along the ground. If not, use Pythagoras to find this horizontal distance.

    Quote Originally Posted by BrettSGLane View Post
    The ball falls with velocity 32 ft/sec
    Acceleration, surely?

    Quote Originally Posted by BrettSGLane View Post
    and the shadow is cast by the sun.
    So you can take the angle of its rays as constant.

    So vertical and horizontal distances covered, and therefore velocities, will remain in proportion.

    And the proportion is 5:1, or else root 24 to one, on the alternative interpretation of 'shadow is 500 ft from ball'.

    Use v^2 = u^2 + 2as.
    Last edited by tom@ballooncalculus; July 5th 2012 at 11:52 AM.
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    Re: Speed of shadow.

    The problem is exactly how it's phrased in my book.

    The way I see it is, you can make a right triangle from the initial position of the ball, the initial position of the shadow and the point on the ground where the ball will land. Are you saying the angles within that triangle are constant, cause to me they're not,though I could be visualizing it wrong.
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    Re: Speed of shadow.

    Quote Originally Posted by BrettSGLane View Post
    The problem is exactly how it's phrased in my book.
    Word for word? What about 'falls with velocity 32 m/s' ? Not acceleration? If initial velocity, why not say? ...

    Anyway,

    Quote Originally Posted by BrettSGLane View Post
    The way I see it is, you can make a right triangle from the initial position of the ball, the initial position of the shadow and the point on the ground where the ball will land.
    Good.

    Quote Originally Posted by BrettSGLane View Post
    Are you saying the angles within that triangle are constant
    Not equal to each other, but equal to their own corresponding counterparts in any smaller triangle later on.

    The sun being so hugely far and high in the sky, we must be expected to assume that the vertical height of the triangle remains one fifth of the horizontal base.
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  6. #6
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    Re: Speed of shadow.

    Yes, that would be my interpretation. Initially, you have a right triangle with the ball at one vertex, the right angle 100 ft below the ball, and the third vertex 500 ft to one side of the right angle. At each point, as the ball falls, the line from the ball to its shadow is parallel to the hypotenuse of the original triangle. By "similar triangles", when the ball is x feet above the ground, taking y as the distance to its shadow, x/y= 100/500.
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    Re: Speed of shadow.

    Well, if the angles are constant then the answer is -64(6)^(1/2).
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    Re: Speed of shadow.

    Drop the minus sign as direction doesn't matter.

    Now, where's the 6 from?

    Are you using v^2 = u^2 + 2as ...?

    And when you've got the vertical speed, scale appropriately (i.e. times 5 or, just conceivably, root 24).
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    Re: Speed of shadow.

    Square root of 24 is equal to 2 times the square root of 6.
    I'm assuming that the ball is falling at a constant velocity of 32 ft/sec. 32*2(6)^(1/2) is equal to 64(6)^(1/2).
    What formula are you referring to cause I don't recognize it.
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  10. #10
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    Re: Speed of shadow.

    Quote Originally Posted by BrettSGLane View Post
    Square root of 24 is equal to 2 times the square root of 6.
    Oh, fair enough. But do this:

    v^2 = u^2 + 2as

    first. (Or be careful.) Where v is final velocity (of the ball when it reaches ground level), u = initial velocity = 0, a = acceleration = 32 m/s^2, and s = distance = 100.

    Equations of motion - Wikipedia, the free encyclopedia #SUVAT_equations

    Quote Originally Posted by BrettSGLane View Post
    I'm assuming that the ball is falling at a constant velocity of 32 ft/sec.
    One is bound to ask, "on which planet?!"
    Last edited by tom@ballooncalculus; July 5th 2012 at 01:54 PM.
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    Re: Speed of shadow.

    Forgot all about that equation, I calculated the time first and then used it to calculate the speed. That wouldn't have been necessary with your equation.
    In this case I am assuming that the ball is accelerating at 32 ft/s^2.
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