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Thread: [SOLVED] SAT math page 657 #18

  1. #1
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    [SOLVED] SAT math page 657 #18

    h(t)=c-(d-4t)^2

    At time t=0, a ball was thrown upward from an initial height of 6 feet. Until the ball hit the ground, its height, in feet, after t seconds was given by the function h above, in which c and d are positive constants. If the ball reached its maximum height of 106 feet at the time t=2.5, what was the height, in feet, of the ball at time t=1?

    Thanks in advance!!

    The correct answer is 70 but why?
    Last edited by fabxx; Sep 28th 2008 at 03:30 AM.
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  2. #2
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    Hi fabxx,

    We have $\displaystyle h(t)=c-(d-4t)^2$ . Note that in the text it states, "at time $\displaystyle t=0$, a ball was thrown upward from an initial height of $\displaystyle 6$ feet" so we can let $\displaystyle t=0$ and $\displaystyle h=6$ . This yields $\displaystyle h(1)=c-d^2=6$ . Call this equation $\displaystyle (*)$ . Also note in the text; "$\displaystyle 106$ feet at the time $\displaystyle t=2.5$" thus letting $\displaystyle t=\frac{5}{2}$ and $\displaystyle h=106 $ we have $\displaystyle h\left(\frac{5}{2}\right)=c-\left[d-4\left(\frac{5}{2}\right)\right]^2=c-(d-10)^2=106$ . Define this as equation $\displaystyle (**) $.

    Now we have two equations with two unknowns, namely $\displaystyle c$ and $\displaystyle d$;

    $\displaystyle c-d^2=6~~~(*)$

    $\displaystyle c-(d-10)^2=106~~~(**)$

    All that remains is to solve these two equations simultaneously finding $\displaystyle c$ and $\displaystyle d$ then subbing in $\displaystyle t=1$ and the values for $\displaystyle c$ and $\displaystyle d$. Can you continue?
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  3. #3
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    I got d=10 and c=106 and substitute t=1 I got 70 as the final answer!

    Thanks Sean12345!
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