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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 (in seconds). As an improper fraction, for how long is the cannonball above a height of $5$ meters?

Oct 19, 2021

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For how long is the cannonball above a height of 5 meters?

Hello Guest!

$$h(t) = -4.9t^2 + 14t - 0.4$$

$$-4.9t^2 + 14t - 0.4=5\\ -4.9t^2 + 14t - 5.4=0$$

$$x = {-b \pm \sqrt{b^2-4ac} \over 2a}$$

$$t= {-14 \pm \sqrt{196-4\cdot 4.9\cdot 0.4} \over 2\cdot (-4.9)}\\ t=\dfrac{-14\pm 13.7171}{-9.8}$$

$$t\in \{0.02886s,2.8283s\}$$

2.8283s - 0.0288s = 2.799s long is the cannonball above a height of 5 meters.

!

Oct 19, 2021