The arrow is at a height of 48 ft after approximately 0.55 seconds and after 5.45 seconds.
Explanation
The given formula is: [tex]s=V_{0}t-16t^2[/tex]
If the initial velocity is 96 ft/s , that means [tex]V_{0}=96[/tex]
For finding the time the arrow takes to reach a height of 48 ft, we will plug [tex]s= 48[/tex] into the above formula. So......
[tex]48=96t-16t^2\\ \\ 16t^2-96t+48=0\\ \\ 16(t^2-6t+3)=0\\ \\ t^2-6t+3=0\\ \\ t^2-6t =-3\\ \\ t^2-6t+9=-3+9\\ \\ (t-3)^2 = 6\\ \\ t-3= \pm \sqrt{6} \\ \\ t=3\pm \sqrt{6}\\ \\ t = 5.45 , 0.55[/tex]
So, the arrow is at a height of 48 ft after approximately 0.55 seconds and after 5.45 seconds.
