Answer: 3 seconds
We are given the following function:
[tex]h(t)=-16{t}^{2}+V.t+h(o)[/tex] (1)
Where:
[tex]t[/tex] is the time the ball is in the air
[tex]V=47ft/s[/tex] the initial upward velocity
[tex]h(o)=3ft[/tex] the initial height of the ball
[tex]h(t)[/tex] is the final height of the ball. If no one catches it, this will be zero
So, equation (1) changes to:
[tex]-16{t}^{2}+V.t+h(o)=0[/tex] (2)
Substituting the known values:
[tex]-16{t}^{2}+(47ft/s).t+3ft=0[/tex] (3)
This is a quadratic equation in the form [tex]a{t}^{2}+b.t+c=0[/tex]. In order to find [tex]t[/tex] we can use the quadratic formula for the roots:
[tex]t=\frac{-b\pm\sqrt{-4ac}}{2a}[/tex] (4)
Where [tex]a=-16[/tex], [tex]b=47[/tex] and [tex]c=3[/tex]
Substituting this values in (4):
[tex]t=\frac{-47\pm\sqrt{-4(-16)(3)}}{2(-16)}[/tex] (5)
[tex]t=\frac{-47\pm\sqrt{2401}}{-32}[/tex] (6)
[tex]t=\frac{-47\pm\49}{-32}[/tex] (7)
For [tex]t1[/tex]:
[tex]t1=\frac{-47+49}{-32}=-0.0625 s[/tex] (8)>>>> This result does not work for us because is negative
For [tex]t2[/tex]:
[tex]t2=\frac{-47-49}{-32}=3 s[/tex] (9)>>>This is the result
Therefore:
If no one catches the ball, it will be 3 s in the air
