The height (in meters) of a shot cannonball follows a trajectory given by h(t) = -4.9t^2 + 14t - 0.4 at time t (in seconds). For how many seconds is the height of the cannonball at least \(8\) meters?
+130020 We want to solve this :
-4.9t^2 + 4t - .4 ≥ 8 subtract 8 from both sides
-4.9t^2 + 4t - 8.4 ≥ 0 multiply through by -1
4.9t^2 - 4t + 8.4 ≥ 0
Using x for t, look at the graph here : https://www.desmos.com/calculator/g6yr1y8a7p
Notice that the height is ≥ 8m between ≈ .117 sec and ≈ .7 sec
+37157 h = 8
8 = -4.9t^2 + 14 t - .4
0 = -4.9t^2 + 14t - 8.4 Use quadratic formula to solve for t the difference in the two t's is the time above 8 m
a = - 4.9 b = 14 c = -8.4
\(t = {-14 \pm \sqrt{14^2-4(-4.9)(-8.4)} \over 2(-4.9)}\) reults in t = 2 and .8571 difference = 1.143 sec <==== time = or above 8 m
