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the height of an arrow shot upward can be given by the formula s = v₀t …

Question

the height of an arrow shot upward can be given by the formula s = v₀t - 16t², where v₀ 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 - 16t². solve 16t² - 96t + 48 = 0. complete the square to write 16t² - 96t + 48 = 0 as √(t - 3)² = 6√. complete solve (t - 3)² = 6. the arrow is at a height of 48 ft after approximately s and after s. done

Explanation:

Step1: Solve the equation $(t - 3)^2=6$

Take the square - root of both sides: $t - 3=\pm\sqrt{6}$.

Step2: Solve for $t$ when $t - 3=\sqrt{6}$

Add 3 to both sides: $t_1=3+\sqrt{6}\approx3 + 2.45=5.45$.

Step3: Solve for $t$ when $t - 3=-\sqrt{6}$

Add 3 to both sides: $t_2=3-\sqrt{6}\approx3 - 2.45 = 0.55$.

Answer:

$0.55$; $5.45$