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Math Help - SAT height/time problem

  1. #1
    m58
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    SAT height/time problem

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

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

    Obviously, this is a downward sloping parabola since it has a maximum.

    So far, I've entered:

    106(2.5) = c - (d - 4(2.5))^2

    but I'm not sure how to proceed from there
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  2. #2
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    Re: SAT height/time problem

    You also know that when t=0 h=6. So 6=c-d^2 . So c=6+d^2. Substitute this for c into your equation and solve for d. Then find c from c=6+d^2.
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  3. #3
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    Re: SAT height/time problem

    Hello, m58!

    At time t = 0, a ball was thrown upward from an initial height of 6 feet.
    Until the hall hit the ground, its height (in feet), after t seconds was given by: / h(t) \"=\:c - (d-4t)^2
    where c and d are positive constants.

    If the ball reached its maximum height of 106 ft at t = 2.5,
    what is the height of the ball, in feet, at time t = 1?

    We are told that when t = 0,\:h = 6
    . . c - (d - 0)^2 \:=\:6 \quad\Rightarrow\quad c - d^2 \:=\:6\;\;[1]

    We are told that when t = 2.5,\:h = 106
    . . c - [d - 4(2.5)]^2 \:=\:106 \quad\Rightarrow\quad c - (d-10)^2 \:=\:106\;\;[2]

    Subtract [1] - [2]: . -d^2 +(d-10)^2 + d^2 \:=\:6-106 \quad\Rightarrow\quad -20d + 100 \:=\:-100

    . . . . . . . . . . . . . . -20d \:=\:-200 \quad\Rightarrow\quad d \:=\:10 \quad\Rightarrow\quad c \:=\:106

    Hence: . h(t) \:=\:106 - (10-4t)^2

    Therefore: . h(1) \;=\;106 - (10-4)^2 \;=\;106 - 36 \;=\;70\text{ ft}
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  4. #4
    m58
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    Re: SAT height/time problem

    Biffboy, Soroban, you've explained it well! Thank you very much (sorry for the delayed reply!)
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