The height of a small rock falling from the top of a 124-ft-tall building with an initial downward velocity of - 30 ft/sec is modeled by the equation h (t) = - 16t2 - 30t + 124, where t is the time in seconds. For which interval of time does the rock remain in the air?

Answer: 0≤ t ≤2

Step-by-step explanation:

Hi, to solve this question the equation must be equal to 0 (h=0)

h (t) = - 16t2 - 30t + 124

0 = - 16t2 - 30t + 124

we have to apply the quadratic formula:

For: ax2 + bx + c

a=-16; b=-30; c=124

x = [ - b ± √b²-4ac] / 2a

Replacing with the values given:

x = [ - (-30) ± √ (-30) ²-4 (-16) 124] / 2 (-16)

x = [ 30 ± √900 + 7,936] / -32

x = [ 30± √8,836] / -32

x = [ 30 ± 94] / -32

Positive:

x = [ 30 + 94] / -32 = - 3.875

Negative:

x = [ 30 - 94] / -32 = 2 seconds

Since time can't be negative, the rock remains in the air for 2 seconds.

0≤ t ≤2

Feel free to ask for more if needed or if you did not understand something.