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Math Help - damsel in distress- need help with this questions

  1. #1
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    damsel in distress- need help with this questions

    1.Assuming a constant rate of acceleration due to gravity of 32 feet per second per
    second, findnd the initial velocity needed to launch a projectile from the surface of the
    earth to a height of 900 feet.

    2. Suppose that the probability distribution function for customers waiting to be
    seated at a certain restaurant on a certain evening has the form f(t) = ce^-ct for
    t > 0, where c = 1/10 is the reciprocal of the average waiting time in minutes. Find
    the time T such that exactly 5% of customers wait longer than T minutes to be
    seated. Round your answer to the nearest minute.
    Last edited by iva17; July 5th 2012 at 06:45 AM.
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  2. #2
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    Re: damsel in distress- need help with this questions

    No.1 You have to learn the formulae for these questions
    The one you need is v^2=u^2+2as where u = initial speed v=final speed a=acceleration and s= distance
    For you v=0 s=900 a=-32 (because it is traveling upwards) and u is the question
    So 0=u^2-(2*32*900) u^2=57600 u= 240
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  3. #3
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    Re: damsel in distress- need help with this questions

    You could also do this problem using the fact that if an object has constant acceleration of -32 f/s^2, its velocity, after t seconds, is -32t+ v0, where v0 is the speed at t= 0, and its height is -16t^2+ v0t+ h0, where h0 is the height when t= 0. You can take h0= 0 and you want to find v0 so that the maximum height (this is a quadratic so the maximum height is at the vertex of the parabola- you can find it by completing the square) is 900.

    For the second problem, you want T such that \int_T^\infty f(t)dt= 0.05.
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  4. #4
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    Re: damsel in distress- need help with this questions

    Hello, iva17!

    1. Assuming a constant rate of acceleration due to gravity of 32 ft/sec2,
    find the initial velocity needed to launch a projectile from the surface of the earth to a height of 900 feet.

    The height y of the projectile is given by: . y \;=\;h_o + v_ot - \tfrac{1}{2}gt^2

    . . where: . \begin{Bmatrix}h_o &=& \text{initial height} \\ v_o &=& \text{initial velocity} \\ g &=& \text{accel'n due to gravity} \end{Bmatrix}

    We are given: . \begin{Bmatrix}h_o &=& 0 \\ g &=& 32\end{Bmatrix}

    The formula becomes: . y \;=\; v_ot - 16t^2

    Assuming we want the minimum initial velocity to reach a height of 900 feet,
    . . the vertex of the parabola is: . t\;=\;\frac{\text{-}b}{2a} \:=\:\frac{\text{-}v_o}{2(\text{-}16)} \:=\:\frac{v_o}{32}

    The formula becomes: . v_o\left(\frac{v_o}{32}\right) - 16\left(\frac{v_o}{32}\right)^2 \;=\;900 \quad\Rightarrow\quad \frac{v_o^2}{64} \:=\:900

    . . . . . v_o^2 \:=\:64\cdot900 \quad\Rightarrow\quad v_o \:=\:8\cdot30


    Therefore: . v_o \:=\: 240\text{ ft/sec}
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