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A firecracker shoots up from a hill 140 feet high with an initial speed of 100 feet per second. Using the formula H(t) = −16t2 + vt + s, approximately how long will it take the firecracker to hit the ground?


5 seconds


7 seconds


9 seconds


11 seconds

Verified Answer :

nobillionaireNobley

H(t) = -16t^2 + vt + s
0 = -16t^2 + 100t + 140
4t^2 - 25t - 35 = 0
t = 7 seconds.

Answer:

7 seconds

Step-by-step explanation:

A firecracker shoots up from a hill 140 feet high with an initial speed of 100 feet per second. Using the formula H(t) = −16t^2 + vt + s

v is the initial speed and s is the initial height

Initial speed v= 100 and initial height = 140

So the equation becomes  H(t) = −16t^2 + 100t + 140

When the firecracket hit the ground the height becomes 0

So we plug in H(t) for 0 and solve for t

\(0 = -16t^2 + 100t + 140\)

Apply quadratic formula

\(x=\frac{-b+-\sqrt{b^2-4ac}}{2a}\)

a= -16, b= 100, c= 140

\(t=\frac{-100+-\sqrt{100^2-4(-16)(140)}}{2(-16)}\)

\(t=\frac{-100+-\sqrt{18960}}{-32}\)

\(t=\frac{-100+-\4sqrt{1185}}{-32}\)

\(t=\frac{25+-\sqrt{1185}}{8}\)

t=-1.18  or t= 7.43

it take 7 seconds for  the firecracker to hit the ground

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