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) …

Question

asked Feb 14, 2020 93.5k views

1 Answer

Eric Schmidt · Answer 1 · 2020-02-18T08:40:55+0000

Answer: 3 seconds

We are given the following function:

$h(t)=-16{t}^(2)+V.t+h(o)$ (1)

Where:

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
we can use the quadratic formula for the roots:

$t=(-b\pm√(-4ac))/(2a)$ (4)

Where
a=-16 ,
b=47 and
c=3

Substituting this values in (4):

$t=(-47\pm√(-4(-16)(3)))/(2(-16))$ (5)

$t=(-47\pm√(2401))/(-32)$ (6)

$t=(-47\pm\49)/(-32)$ (7)

For
:

t1=(-47+49)/(-32)=-0.0625 s (8)>>>> This result does not work for us because is negative

For
:

t2=(-47-49)/(-32)=3 s (9)>>>This is the result

Therefore: