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

t=0.55 sec

step-by-step explanation :

we know that

the measurement of the outer angle is the semi-difference of the arcs which comprises

in this problem

m< xzy is the outer angle

substitute the values

the measure of angle xoy is equal to

> by central angle

substitute the value