Answer:
It takes 7.5 seconds the object to hit the ground.
Step-by-step explanation:
An object is propelled upward from the top of a 300 foot building. The path that the object takes as it falls to the ground can be modeled by:
h = -16t^2 + 80t + 300
where:
t is the time (in seconds) and
h is the corresponding height (in feet) of the object.
How long does it take the object to hit the ground?
When the objet hit the ground:
h=0
Then, equaling h (the equation) to zero:
[tex]-16t^{2}+80t+300=0[/tex]
This is a quadratic equation. Using the quadratic formula:
[tex]at^2+bt+c=0; a=-16, b=80, c=300[/tex]
[tex]t=\frac{-b+-\sqrt{b^2-4ac} }{2a}\\ t=\frac{-80+-\sqrt{80^2-4(-16)(300)} }{2(-16)}\\ t=\frac{-80+-\sqrt{6,400+19,200} }{-32}\\ t=\frac{-80+-\sqrt{25,600} }{-32}\\ t=\frac{-80+-160}{-32}[/tex]
Two solutions:
[tex]\left \{ {{t_{1} =\frac{-80+160}{-32} } \atop {t_{2} =\frac{-80-160}{-32} }} \right.[/tex]
[tex]\left \{ {{t_{1} =\frac{80}{-32} } \atop {t_{2} =\frac{-240}{-32} }} \right.[/tex]
[tex]\left \{ {{t_{1} =-2.5} \atop {t_{2} =7.5}} \right.[/tex]
The first solution is not possible, because the time can't be a negative number, then the solution is the second one: t=7.5 seconds
Answer: It takes 7.5 seconds the object to hit the ground.
