Question

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

Answer 1

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

Answer 2

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