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Pre-Calc

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The height s of a ball, in feet, thrown with an initial velocity of 112 feet per second from an initial height of 20 feet is given as a function of time t (in seconds) by
s(t)= -16t^2+112t+20. What is the maximum height of the ball? At what time does the maximum height occur?

I know how to find the maximum height. You find the vertex which is 3.5 then plug it back into the original equation.

Now, when it says what time, I am not sure if it wants the vertex or should I use the quadratic to solve it?

Please help.

  • Pre-Calc - ,

    The maximum height is the vertex.

    This would be an application of completing the squares to convert the function into the standard form:
    s(t)=a(t-h)²+k
    where t=h at the vertex (3.5) and
    k=maximum height (216 above datum)

  • Pre-Calc - ,

    Ok thanks.

  • Pre-Calc - ,

    You're welcome!

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