Question

A football player kicks a ball with an initial upward velocity of 47 ft./s. The initial height of the ball is 3 feet. The function h(t)=-16t^2+vt+h0 models the height (in feet) of the ball, where V is the initial upward velocity and h0 is the initial height. If no one catches the ball, how long will it be in the air?

Answer 1

Answer: 3 seconds

We are given the following function:

$h(t)=-16{t}^{2}+V.t+h(o)$    (1)

Where:

$t$ is the time the ball is in the air

$V=47ft/s$ the initial upward velocity

$h(o)=3ft$ the initial height of the ball

$h(t)$ is the final height of the ball. If no one catches it, this will be zero

So, equation (1) changes to:

$-16{t}^{2}+V.t+h(o)=0$    (2)

Substituting the known values:

$-16{t}^{2}+(47ft/s).t+3ft=0$    (3)

This is a quadratic equation in the form $a{t}^{2}+b.t+c=0$. In order to find $t$ we can use the quadratic formula for the roots:

$t=\frac{-b\pm\sqrt{-4ac}}{2a}$    (4)

Where $a=-16$, $b=47$ and  $c=3$

Substituting this values in (4):

$t=\frac{-47\pm\sqrt{-4(-16)(3)}}{2(-16)}$    (5)

$t=\frac{-47\pm\sqrt{2401}}{-32}$    (6)

$t=\frac{-47\pm\49}{-32}$    (7)

For $t1$:

$t1=\frac{-47+49}{-32}=-0.0625 s$      (8)>>>> This result does not work for us because is negative

For $t2$:

$t2=\frac{-47-49}{-32}=3 s$   (9)>>>This is the result

Therefore:

If no one catches the ball, it will be 3 s in the air