Question

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?

Answer 1

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:

$x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$

On comparing the given equation with [1] we have;  

a = -16 ,b = 80 and c = 300

then;

$t= \frac{-80\pm \sqrt{(80)^2-4(-16)(300)}}{2(-16)}$

$t= \frac{-80\pm \sqrt{6400+19200}}{-32}$

$t= \frac{-80\pm \sqrt{25600}}{-32}$  

Simplify:

$t = -\frac{5}{2}$ = -2.5 sec and $t = \frac{15}{2}$ = 7.5 sec

Time can't be in negative;

therefore, the time it took the object to hit the ground is 7.5 sec

Answer 2

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:

$-16t^{2}+80t+300=0$

This is a quadratic equation. Using the quadratic formula:

$at^2+bt+c=0; a=-16, b=80, c=300$

$t=\frac{-b+-\sqrt{b^2-4ac} }{2a}\\ t=\frac{-80+-\sqrt{80^2-4(-16)(300)} }{2(-16)}\\ t=\frac{-80+-\sqrt{6,400+19,200} }{-32}\\ t=\frac{-80+-\sqrt{25,600} }{-32}\\ t=\frac{-80+-160}{-32}$

Two solutions:

$\left \{ {{t_{1} =\frac{-80+160}{-32} } \atop {t_{2} =\frac{-80-160}{-32} }} \right.$

$\left \{ {{t_{1} =\frac{80}{-32} } \atop {t_{2} =\frac{-240}{-32} }} \right.$

$\left \{ {{t_{1} =-2.5} \atop {t_{2} =7.5}} \right.$

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.