Question

A juggler tosses a ring into the air. The height of the ring in feet above the juggler’s hands can be modeled by the function f (x) = -16 x 2 + 16x, where x is the time in seconds after the ring is tossed. Find the ring’s maximum height above the juggler’s hands and the time it takes the ring to reach this height. Then find how long the ring is in the air.

Answer 1

To find the maximum height of the ring, we write the function in vertex form,

Here we go. We were given;

$f(x)=-16x^2+16x$

We factor  $-16$ to obtain,

$f(x)=-16(x^2-x)$

We now complete the square within the parenthesis so we add and subtract $-16(\frac{-1}{2})^2$ to obtain.

$f(x)=-16(x^2-x+(\frac{-1}{2})^2) +16(\frac{-1}{2})^2$

Taking note of the perfect square in the parethesis, we simplify to get,

$f(x)=-16(x-\frac{1}{2})^2 +16\times (\frac{1}{4})$

This further gives

$f(x)=-16(x-\frac{1}{2})^2 +4$

The vertex of the above parabola is $(\frac{1}{2},4)$.

The y-value of the vertex of the parabola gives us the maximum height of the ring above the juggler's hand.

Therefore the maximum height of the ring is $4$ feet.

How long the ring is in the air can be determined by finding the time the ring left the juggler's hand and subtracting it from the time it came back into his hands. This will be the difference between the root of the corresponding quadratic equation.

We set the equation to zero now and solve.

$-16(x-\frac{1}{2})^2 +4=0$

$-16(x-\frac{1}{2})^2 =-4$

$(x-\frac{1}{2})^2 =\frac{1}{4}$

$x-\frac{1}{2} =\pm \sqrt{\frac{1}{4}}$

$x =-\frac{1}{2} \pm \frac{1}{2}$

$x =\frac{1}{2} \pm \frac{1}{2}$

$x =0 \: or\: x=1$

Hence the ball was in the air for $x =1-0=1$ seconds.

See diagram