91Ó°ÊÓ

If $a{x}^2+bx+c=0$, with $a≠0$, then $x=\frac{−bÂ\pm \sqrt{b^2−4ac}}{2a}$. Solving using factoring can be done by splitting the middle term of the equation and then making groups and equating them to zero.

The ball is no longer in the air when it lands on the ground which occurs when $h=0$, so substitute 0 for h into the equation $h=−16t^2+60t+3$.

$\begin{array}{l}h=−16t^2+60t+3\\ 0=−16t^2+60t+3\end{array}$

Solve for t by using the quadratic formula. Substitute $-16$ for a, 60 for b, and 3 for c into the equation $t=\frac{−bÂ\pm \sqrt{b^2−4ac}}{2a}$.

$$\begin{gathered} t=\frac{−60Â\pm \sqrt{{\left(60\right)}^2−4\left(−16\right)\text{ }\left(3\right)}}{2\left(−16\right)} \\ =\frac{−\text{ â¶Ä‰}60\text{ â¶Ä‰}Â\pm \text{ â¶Ä‰}\sqrt{3600+192}}{−32} \\ =\frac{−60Â\pm \sqrt{3792}}{−32} \\ t=\frac{−60−\sqrt{3792}}{−32}\text{ â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰and â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰}\frac{−\text{ â¶Ä‰}60\text{ â¶Ä‰}+\text{ â¶Ä‰}\sqrt{3792}}{−32} \\ t≈\text{ â¶Ä‰}0\text{ â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰and â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰}t≈3.8 \end{gathered}$$

Time $t≈0$second corresponds to when the ball is hit, so $\text{ }t≈3.8$seconds corresponds to when it lands on the ground. Therefore, the ball is then in the air for about 3.8 seconds.