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

Answer 1

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

Explanation

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

If the initial velocity is 96 ft/s , that means  $V_{0}=96$

For finding the time the arrow takes to reach a height of 48 ft, we will plug $s= 48$ into the above formula. So......

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

So, the arrow is at a height of 48 ft after approximately 0.55 seconds and after 5.45 seconds.

Answer 2

The given values are:

s= 48 ft

v0= 96 ft/s

t= unknown

We are asked to find for t:

48 = 96 t – 16t2

Rearranging the equation:

16t² - 96t + 48 = 0

Dividing whole equation by 16:

t² - 6t + 3 = 0

By completing the squares and factoring:

t² - 6t + 3 + 6 = 6

(t - 3)² = 6

t -3 = +/- 2.45

t = 5.45s , 0.55s

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