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. H

Question

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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. H

ow 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 . Solve (t - 3)² = 6. The arrow is at a height of 48 ft after approximately s and after s.

Mathematics

2 answers:

Svet_ta [4.3K] · Answer 1 · 2026-02-20T14:25:21+03:00

The arrow reaches a height of 48 feet at about 0.55 seconds and again at approximately 5.45 seconds.

Explanation

The formula provided is: $s=V_{0}t-16t^2$

Given the initial velocity as 96 ft/s, we have $V_{0}=96$

To determine when the arrow attains a height of 48 feet, substitute $s= 48$ into the formula:

$48=96t-16t^2\\ \\ 16t^2-96t+48=0\\ \\ 16(t^2-6t+3)=0\\ \\ t^2-6t+3=0\\ \\ t^2-6t =-3\\ \\ t^2-6t+9=-3+9\\ \\ (t-3)^2 = 6\\ \\ t-3= \pm \sqrt{6} \\ \\ t=3\pm \sqrt{6}\\ \\ t = 5.45 , 0.55$