Question

the height of an arrow shot upward can be given by the formula ( s = v_0t - 16t^2 ), where ( v_0 ) is the initial velocity and ( t ) is time. how long does it take for the arrow to reach a height of 48 ft if it has an initial velocity of 96 ft/s? round to the nearest hundredth.
the equation that represents the problem is ( 48 = 96t - 16t^2 ).
solve ( 16t^2 - 96t + 48 = 0 ).
complete the square to write ( 16t^2 - 96t + 48 = 0 ) as ( (t - 3)^2 = 6 ).
solve ( (t - 3)^2 = 6 ). the arrow is at a height of 48 ft after approximately ( square ) s and after ( square ) s.

Explanation:

Step1: Take square root of both sides

Given \((t - 3)^2=6\), take the square root of both sides: \(t - 3=\pm\sqrt{6}\)

Step2: Solve for t

Add 3 to both sides: \(t = 3\pm\sqrt{6}\)
Calculate the numerical values: \(\sqrt{6}\approx2.45\)
So \(t_1 = 3 + 2.45 = 5.45\) and \(t_2 = 3 - 2.45 = 0.55\)

Answer:

The arrow is at a height of 48 ft after approximately 0.55 s and 5.45 s.