Answer
$8.8$ seconds
Work Step by Step
The ball hits the ground when $s(t)=0$.
We have to solve the quadratic equation:
$$-16t^2+140t+3=0.$$
The equation is in the standard form.
To solve the equation $ax^2+bx+c=0$ we will use the quadratic formula:
$$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$$
Identify $a$, $b$, $c$:
$$\begin{align*}
a&=-16\\
b&=140\\
c&=3.
\end{align*}$$
We solve the given equation by substituting the values of $a$, $b$, $c$ in the quadratic formula:
$$\begin{align*}
x&=\dfrac{-140\pm\sqrt{140^2-4(-16)(3)}}{2(-16))}\\
&\approx \dfrac{-140\pm 140.68}{-32}\\
x_1&=\dfrac{-140-140.68}{-32}\approx 8,8\\
x_2&=\dfrac{-140+140.68}{-32}\approx -0.02.\\
\end{align*}$$
Because $t>0$ the only solution is $t\approx 8.8$ seconds.
