Question

An arrow is shot straight up from the roof of a 60-ft tall building with an initial velocity of 68 ft/sec. The height of the arrow is given by:

h(t) = -16t² + 68t + 60
where h(t) is the height, in feet, and t is time, in seconds. At what time or times will the arrow be 112 feet off the ground? Round your answer to the hundredths place, if necessary.
A. t = 1.43 sec
B. t = 2.86 sec
C. t = 3.50 sec
D. t = 4.67 sec

Answer 1

The arrow will be 112 feet off the ground at times t = 1.43 sec and t = 4.67 sec.

Step-by-step explanation:

To find the time(s) when the arrow will be 112 feet off the ground, we can set up an equation using the given height function h(t). We set h(t) equal to 112 and solve for t.

-16t² + 68t + 60 = 112

-16t² + 68t - 52 = 0

To solve this quadratic equation, we can use the quadratic formula:

t = (-b ± sqrt(b² - 4ac))/(2a)

Plugging in the values from our equation:

t = (-68 ± sqrt((68)² - 4(-16)(-52))) / (2(-16))

After simplifying, we get t = 1.43 sec and t = 4.67 sec.

Therefore, the arrow will be 112 feet off the ground at times t = 1.43 sec (Option A) and t = 4.67 sec (Option D).