t = 1.33 sec
Solution:
Given data:
Velocity [tex](v_0)[/tex] = 17.5 ft/s
Height [tex](h_0)[/tex] = 5 ft
The height can be modeled by a quadratic equation
[tex]h(t)=-16t^2+v_0t+h_0[/tex]
where h is the height and t is the time
[tex]h(t)=-16t^2+17.5t+5[/tex]
[tex]-16t^2+17.5t+5=0[/tex]
a = –16, b = 17.5, c = 5
It looks like a quadratic equation. we can solve it by quadratic formula,
[tex]$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}[/tex]
[tex]$\Rightarrow t=\frac{-17.5 \pm \sqrt{(-17.5)^{2}-4\times (-16)(5)}}{2 (-16)}[/tex]
[tex]$\Rightarrow t= \frac{-17.5 \pm \sqrt{306.25+ 320}}{-32}[/tex]
[tex]$\Rightarrow t= \frac{-17.5 \pm 25.025}{-32}[/tex]
[tex]$\Rightarrow t= \frac{-17.5 - 25.025}{-32}, \ t= \frac{-17.5 + 25.025}{-32}[/tex]
[tex]$\Rightarrow t= 1.33, \ t= -0.24[/tex]
Time cannot be in negative. So neglect t = –0.235.
t = 1.33 sec
Hence Amy have to react 1.33 sec before the volleyball hits the ground.