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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?

Feb 14, 2022

#1
+122390
0

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

Feb 14, 2022
#2
+36417
+1

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

Feb 14, 2022