The height of a rock thrown off a cliff is given by h(t) = -2t2 + 12t + 14. When does this rock hit the ground?
Thank you!
-2t^2 + 12t + 14. =0
-2(t^2-6t-7) =0
-2(t-7)(t+1) =0
t-7 =0
t = 7
t+1 =0
t = -1
hit ground 7
To find when the rock hits the ground, we need to determine the value of t when the height, h(t), is equal to zero. So we can set up the equation:
-2t² + 12t + 14 = 0
To solve this quadratic equation, we can use the quadratic formula:
t = (-b ± √(b² - 4ac)) / 2a
In this equation, a = -2, b = 12, and c = 14. Substituting the values into the quadratic formula:
t = (-(12) ± √((12)² - 4(-2)(14))) / (2(-2))
Simplifying further:
t = (-12 ± √(144 + 112)) / -4
t = (-12 ± √(256)) / -4
t = (-12 ± 16) / -4
Using the positive square root:
t = (-12 + 16) / -4
t = 4 / -4
t = -1
Using the negative square root:
t = (-12 - 16) / -4
t = -28 / -4
t = 7
So the rock hits the ground at two different times, t = -1 and t = 7. However, since time cannot be negative, we can conclude that the rock hits the ground at t = 7 seconds.