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

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