Example8.4.5

A physics class launches a tennis ball from a rooftop that is \(90.2\) feet above the ground. They fire it directly upward at a speed of \(14.4\) feet per second and measure the time it takes for the ball to hit the ground below. We can model the height of the tennis ball, \(h\text{,}\) in feet, with the quadratic equation \(h=-16x^2+14.4x+90.2\text{,}\) where \(x\) represents the time in seconds after the launch. According to the model, when should the ball hit the ground? Round the time to one decimal place.

The ground has a height of \(0\) feet. Substituting \(0\) for \(h\) in the equation, we have this quadratic equation:

\begin{equation*} 0=-16x^2+14.4x+90.2 \end{equation*}

We cannot solve this equation with factoring or the square root property, so we will use the quadratic formula. First we will identify that \(\highlight{a=-16}\text{,}\) \(\highlight{b=14.4}\) and \(\highlight{c=90.2}\text{,}\) and substitute them into the formula:

\begin{align*} x\amp=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ x\amp=\frac{-(\substitute{14.4})\pm\sqrt{(\substitute{14.4})^2-4(\substitute{-16})(\substitute{90.2})}}{2(\substitute{-16})}\\ x\amp=\frac{-14.4\pm\sqrt{207.36-(-5772.8)}}{-32}\\ x\amp=\frac{-14.4\pm\sqrt{207.36+5772.8}}{-32}\\ x\amp=\frac{-14.4\pm\sqrt{5980.16}}{-32}\\ \end{align*}

These are the exact solutions but because we have a context we want to approximate the solutions with decimals.

\begin{align*} x\amp\approx-2.0\text{ or }x\approx2.9 \end{align*}

We don't use the negative solution because a negative time does not make sense in this context. The ball will hit the ground approximately \(2.9\) seconds after it is launched.

in-context