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

It takes 7.5 seconds the object to hit the ground.

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=(-b+-√(b^2-4ac) )/(2a)\\ t=(-80+-√(80^2-4(-16)(300)) )/(2(-16))\\ t=(-80+-√(6,400+19,200) )/(-32)\\ t=(-80+-√(25,600) )/(-32)\\ t=(-80+-160)/(-32)`

Two solutions:

`\left \{ {{t_(1) =(-80+160)/(-32) } \atop {t_(2) =(-80-160)/(-32) }} \right.`

`\left \{ {{t_(1) =(80)/(-32) } \atop {t_(2) =(-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.

Answer 2

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 = (-b\pm √(b^2-4ac))/(2a)`

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

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

then;

`t= (-80\pm √((80)^2-4(-16)(300)))/(2(-16))`

`t= (-80\pm √(6400+19200))/(-32)`

`t= (-80\pm √(25600))/(-32)`

Simplify:

`t = -(5)/(2)` = -2.5 sec and
`t = (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