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