Question

the equation (h(t)= - 16t^{2}+40t + 5) gives the height (h(t)), in feet, of a baseball as a function of time (t), in seconds. after it is hit, how long does it take for the baseball to hit the ground to the nearest tenth of a second?

Explanation:

Step1: Set height equal to 0

We want to find when the baseball hits the ground, so we set $h(t)=0$. The equation $h(t)= - 16t^{2}+40t + 5$ becomes $-16t^{2}+40t + 5=0$.

Step2: Use the quadratic formula

The quadratic formula for a quadratic equation $ax^{2}+bx + c = 0$ is $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here, $a=-16$, $b = 40$, and $c = 5$. First, calculate the discriminant $\Delta=b^{2}-4ac=(40)^{2}-4\times(-16)\times5=1600 + 320=1920$.

Step3: Calculate the values of t

$t=\frac{-40\pm\sqrt{1920}}{-32}=\frac{-40\pm8\sqrt{30}}{-32}=\frac{-40\pm8\times5.477}{-32}$. We have two solutions for $t$: $t_1=\frac{-40 + 8\times5.477}{-32}$ and $t_2=\frac{-40-8\times5.477}{-32}$. Since time cannot be negative, we discard the negative - valued solution. $t=\frac{-40 + 43.816}{-32}$ is negative, and $t=\frac{-40-43.816}{-32}=\frac{-83.816}{-32}\approx2.6$.

Answer:

$t\approx2.6$ seconds