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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). As an improper fraction, for how long is the cannonball above a height of 6 meters?

 Aug 3, 2018
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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).

As an improper fraction, for how long is the cannonball above a height of 6 meters?

 

\(\begin{array}{|rcll|} \hline -4.9t^2 + 14t - 0.4 &=& 6 \\ -4.9t^2 + 14t - 0.4 - 6 &=& 0 \quad & | \quad \\ -4.9t^2 + 14t - 6.4 &=& 0 \quad & | \quad \cdot (-1) \\ 4.9t^2 - 14t + 6.4 &=& 0 \\\\ t &=& \dfrac{14\pm \sqrt{14^2-4\cdot 4.9\cdot 6.4 } } { 2\cdot 4.9} \\\\ &=& \dfrac{14\pm \sqrt{196-125.44 } } {9.8} \\\\ &=& \dfrac{14\pm \sqrt{70.56} } {9.8} \\\\ &=& \dfrac{14\pm 8.4 } {9.8} \\\\ t_1 &=& \dfrac{14- 8.4 } {9.8} \\\\ \mathbf{t_1} & \mathbf{=} & \mathbf{ \dfrac{5.6 } {9.8} } \\\\ t_2 &=& \dfrac{14+ 8.4 } {9.8} \\\\ \mathbf{t_2} & \mathbf{=} & \mathbf{\dfrac{22.4} {9.8}} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline t_2-t_1 &=& \dfrac{22.4} {9.8} - \dfrac{5.6 } {9.8} \\\\ &=& \dfrac{22.4-5.6} {9.8} \\\\ &=& \dfrac{16.8} {9.8} \\\\ &=& 1.71428571429\ \text{seconds} \\ \hline \end{array} \)

 

The cannonball is \(\approx 1.7\ \text{seconds} = (\dfrac{16.8} {9.8}\ \text{seconds})\)above a height of 6 meters?

 

 

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 Aug 3, 2018

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