Answer:
As per the statement:
The path that the object takes as it falls to the ground can be modeled by
h =-16t^2 + 80t + 300
where
h is the height of the objects and
t is the time (in seconds)
At t = 0 , h = 300 ft
When the objects hit the ground, h = 0
then;
-16t^2+80t+300=0
For a quadratic equation: ax^2+bx+c=0 ......[1]
the solution for the equation is given by:
[tex]x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
On comparing the given equation with [1] we have;
a = -16 ,b = 80 and c = 300
then;
[tex]t= \frac{-80\pm \sqrt{(80)^2-4(-16)(300)}}{2(-16)}[/tex]
[tex]t= \frac{-80\pm \sqrt{6400+19200}}{-32}[/tex]
[tex]t= \frac{-80\pm \sqrt{25600}}{-32}[/tex]
Simplify:
[tex]t = -\frac{5}{2}[/tex] = -2.5 sec and [tex]t = \frac{15}{2}[/tex] = 7.5 sec
Time can't be in negative;
therefore, the time it took the object to hit the ground is 7.5 sec