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Math Help - speed of a rocket

  1. #1
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    speed of a rocket

    A rocket is rising vertically from a point on the ground 100 meters from an observer at ground level. The observer notes that the angle of elevation is increasing at a rate of 12 degrees per second when the angle of elevation is 60 degrees.

    Find the speed of the rocket at that instant.
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  2. #2
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    Notice when it is at 60 degrees up we can draw a 30, 60, 90 triangle which means the rocket will be 200 meters off the ground at this point

    It took the rocket 5 seconds to get there so use the equation

    r = \frac{d}{t} where 200 is distance and 5 seconds is the amount of time it took.
    Last edited by SnipedYou; October 22nd 2007 at 06:38 PM.
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  3. #3
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    whats the equation
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  4. #4
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    Quote Originally Posted by Mr_Green View Post
    A rocket is rising vertically from a point on the ground 100 meters from an observer at ground level. The observer notes that the angle of elevation is increasing at a rate of 12 degrees per second when the angle of elevation is 60 degrees.

    Find the speed of the rocket at that instant.
    Draw a quick sketch. Call the horizontal distance from the observer to the point where the rocket takes off x (we know this is 100 m), and call y the vertical distance from the ground to the rocket. These are two legs of a right triangle, so the angle between x and the hypotenuse (the line of sight between the observer and the rocket) will be given by
    tan(\theta) = \frac{y}{x}

    So
    y = x~tan(\theta)

    When the angle \theta is 60 degrees we know that the rate the angle is changing is 12 degrees per second. The question is asking, at that point in time, how fast is y changing? The answer is given by the first time derivative of y, which is equal to the vertical speed:
    v = \frac{dy}{dt} = x~sec^2{\theta}~\frac{d\theta}{dt}

    Edit: The rule above only works when \theta is in radians. We need to convert 12 degrees/s into rad/s:
    \frac{d\theta}{dt} = \frac{12^o}{1~s} \cdot \frac{\pi~rad}{180^o} = 0.20944~rad/s

    So when \theta is 60 degrees:
    v = 100 \cdot sec^2(60^o) \cdot 0.20944 = 83.7758~m/s

    -Dan
    Last edited by topsquark; October 22nd 2007 at 07:54 PM.
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    Hey Mr_Green? Do you go to SCN?

    anyways...

    i posted this same question on another forum and they provided me with the answer:

    4189 m/sec

    unfortunately they didn't show any work.

    whos got the correct answer?
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by DINOCALC09 View Post
    Hey Mr_Green? Do you go to SCN?

    anyways...

    i posted this same question on another forum and they provided me with the answer:

    4189 m/sec

    unfortunately they didn't show any work.

    whos got the correct answer?
    This is a "simple" rate of change problem. Until anyone finds a problem in my solution, 4800 m/s should be the correct answer.

    Edit: That's what overconfidence does to a person. I found my own mistake. The correct answer is 83.7758 m/s. I have fixed my original post.

    -Dan
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  7. #7
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    you are a good explainer topS. thankyou very mcuh. i understand.
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