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=(-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`:

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

For
`t2`:

`t2=(-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