• anonymous
Tiffany kicks a soccer ball off the ground and in the air with an initial velocity of 28 feet per second. Using the formula H(t) = -16t2 + vt + s, what is the maximum height the soccer ball reaches
  • Stacey Warren - Expert
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  • jamiebookeater
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  • anonymous
In this problem s=0 because the ball was initially on the ground. The maximum height is when H'(t) = 0.
  • anonymous
In other words, H'(t) = -32t + 33 = 0. Solving gives t = 33/32. Now plug into original equation and you will get the answer.
  • whpalmer4
@mathgeekanqi Not sure how you came up with H'(t) = -32t+33 when v=28... \[s=0,v=28\]\[H(t) = -16t^2+28t+0\]\(H(t)\) is a parabola in form \(y=ax^2+bx+c\) so the vertex is found at \[(x=-\frac{b}{2a}, y=ax^2+bx+c)\]Here we have \(a=-16, b = 28, c = 0\) giving us \(x=\large{-\frac{28}{2*(-16)} = \frac{28}{32} = \frac{7}{8}}\) Plugging \(t=\large{\frac{7}{8}}\) into our formula for \(H(t)\) we get the maximum height reached by the ball. No need for calculus, but I'll admit that I would go the calculus route if available rather than trust my memory for the vertex of a parabola!

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