Question

A firecracker shoots up from a hill 160 feet high, with an initial speed of 90 feet per second. Using the formula H(t) = −16t2 + vt + s, approximately how long will it take the firecracker to hit the ground?
A. Five seconds
B. Six seconds
C.Seven seconds
D.Eight seconds

Answer 1

H(t) = -16t² + vt + s

Where -16 is half of the gravitational constant of almost 32 ft/sec (downward, thus negative),

v is the initial velocity (90), and

s is the starting height (160)

so we have:

H(t) = -16t² + 90t + 160

How long before it hits the ground? Solve for h(t) = 0:

0 = -16t² + 90t + 160

Divide both sides by -2:

0 = 8t² - 45t - 80

Quadratic equation:

t = [ -b ± √(b² - 4ac)] / (2a)

t = [ -(-45) ± √((-45)² - 4(8)(-80))] / (2(8))

t = [ 45 ± √(2025 + 2560)] / 16

t = [ 45 ± √(4585)] / 16

throwing out the negative time:

t = (45 + √4585) / 16

t ≈ 7.04 seconds