The function $h(t) = - 16{t^2} + 80t$ represents the height of the baseball over time. How long do you think the ball will be in the air?

Hint: $h(t) = - 16{t^2} + 80t + 0$ seems a small amount off. This implies we are releasing the ball from a height of zero feet. Then you have to find the value of $t$ when height is again zero feet after releasing .

Complete step-by-step solution:
It is given that the function $h(t) = - 16{t^2} + 80t$ represents the height of the baseball over time.
We are solving for the $t$ when $h(t) = 0$, therefore we are going to set our function $ - 16{t^2} + 80t = 0$
Since this is often a quadratic with the $C$ term missing, $(a{x^2} + bx + c)$
We will solve by factoring.
$ \Rightarrow - 16{t^2} + 80t = 0$
For finding roots of the original equation, we have to use quadratic formula i.e.,
$\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Now identify $a,b,c$ from the original equation given below,
$ \Rightarrow - 16{t^2} + 80t = 0$
$
   \Rightarrow a = - 16 \\
   \Rightarrow b = 80 \\
   \Rightarrow c = 0 \\
 $
Put these values into the formula of finding the roots of quadratic equations,
$\Rightarrow x = \dfrac{{ - 80 \pm \sqrt {{{80}^2} - 80*( - 16)*(0)} }}{{2*( - 16)}}$
After simplifying and by evaluating exponents and square root of the above equation we get the following simplified expression,
$\Rightarrow x = \dfrac{{ - 80 \pm 80}}{{ - 32}}$
To find the roots of the equations , separate the particular equation into its corresponding parts : one part with the plus sign and the other with the minus sign ,
$
   \Rightarrow {x_1} = \dfrac{{ - 80 + 80}}{{ - 32}} \\
  and \\
   \Rightarrow {x_2} = \dfrac{{ - 80 - 80}}{{ - 32}} \\
 $
Simplify and then isolate $x$to find its corresponding solutions!
$
   \Rightarrow {x_1} = 0 \\
  and \\
   \Rightarrow {x_2} = 5 \\
 $
So , the ball is on the bottom a time , $t = 0$ (right before you throw it) and goes up and comes backpedal and hits the bottom $5$ seconds later, $t = 5$ .

Therefore , the ball will be in the air for 5 second.

Note: The ball is on the bottom a time , $t = 0$ (right before you threw it) . This implies we are releasing the ball from a height of zero feet and goes up and comes backpedal and hits the bottom $5$ seconds later, $t = 5$ .