Question

Se lanza un objeto desde una plataforma.
Su altura (en metros), x segundos después del lanzamiento, está modelada por:
h(x)=-5x^2+20x+60h(x)

¿Cuántos segundos después del lanzamiento el objeto llega al suelo?

Answer 1

Answer:

6 seconds

Step-by-step explanation:

The question in English is

An object is thrown from a platform.

Its height (in meters), x seconds after launch, is modeled by:

h(x)=-5x^2+20x+60

How many seconds after the launch does the object reach the ground?

Let

x ----> the time in seconds

h(x) ---> the height of the object

we have

$h(x)=-5x^2+20x+60$

we know that

When the object hit the ground the height is equal to zero

so

For h(x)=0

we have

$-5x^2+20x+60=60$

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

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

in this problem we have

$-5x^2+20x+60=60$

so

$a=-5\\b=20\\c=60$

substitute in the formula

$x=\frac{-20\pm\sqrt{20^{2}-4(-5)(60)}} {2(-5)}$

$x=\frac{-20\pm\sqrt{1,600}} {-10}$

$x=\frac{-20\pm40} {-10}$

$x=\frac{-20+40} {-10}=-2$

$x=\frac{-20-40} {-10}=6$

The solution is x=6 sec

The he object reach the ground at x=6 seconds