The height of an arrow shot upward can be given by the formula s = vnt - 16, where vo 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 - 1612.

Solve 16t - 960 + 48 = 0.

Complete the square to write 161-961 + 48 = 0 as

I'm going to take a guess that the asker meant

s = v₀t - 16t²

v₀=96 ft/sec

s=48 ft

48 = 96t - 16t²

16t² - 96t + 48 = 0

Cancel a factor of 16.

t² - 6t + 3 = 0

Completing the square,

(t - 3) ² - 9 + 3 = 0

(t - 3) ² = 6

t - 3 = ±√6

t = 3±√6

Round? I hate ruining a nice exact answer. Both ts are positive, one on the way up, one one the way down.

t≈0.55 sec or t≈5.45 sec

How long sounds like we're asking how long is the going up part, which is

Answer: t=0.55 sec