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# Help plz

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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?

Guest Feb 21, 2017
edited by Guest  Feb 21, 2017

#3
+18625
+10

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?

$$\begin{array}{|rcll|} \hline h(t) = -4.9t^2 +14t -0.4 & \ge & 6 \\\\ -4.9t^2 +14t -0.4 & = & 6 \\ -4.9t^2 +14t -0.4 -6 & = & 0 \\ -4.9t^2 +14t -6.4 & = & 0 \\ t &=& \frac{-14\pm \sqrt{ 14^2-4\cdot (-4.9)\cdot (-6.4) }}{2\cdot(-4.9)} \\ t &=& \frac{-14\pm \sqrt{ 196-125.44 }}{-9.8} \\ t &=& \frac{-14\pm \sqrt{ 70.56 }}{-9.8} \\ t &=& \frac{-14\pm 8.4 }{-9.8} \\ t &=& \frac{14\pm 8.4 }{9.8} \\\\ \Delta t &=& t_2-t_1 \\ \Delta t &=& \frac{14 + 8.4 }{9.8}-\left(\frac{14- 8.4 }{9.8} \right) \\ \Delta t &=& \frac{14 + 8.4 -14+8.4}{9.8} \\ \Delta t &=& \frac{2\cdot 8.4}{9.8} \\ \Delta t &=& \frac{8.4}{4.9} \\ \Delta t &=& \frac{84}{49} \\ \Delta t &=& \frac{7\cdot 12}{7\cdot 7} \\ \mathbf{ \Delta t } & \mathbf{=} & \mathbf{\frac{ 12}{ 7}} \\ \hline \end{array}$$

The cannonball is $$\frac{12}{7}$$ seconds above a height of 6 meters

heureka  Feb 21, 2017
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#1
+77073
0

See the graph here :   https://www.desmos.com/calculator/3ceh7ii7po

The  cannonball is above 6m   from ≈ .571s  to ≈ 2.286s

So    2.286 - .571   ≈   343 / 200  seconds    [1.715 sec ]

CPhill  Feb 21, 2017
#2
0

If you solve 6 = -4.9t^2 +14t -0.4 for t, you get that t = 4/7 and t = 16/7. The difference of these is 12/7, so the exact time is 12/7 seconds, as an improper fraction.

Guest Feb 21, 2017
#3
+18625
+10

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?

$$\begin{array}{|rcll|} \hline h(t) = -4.9t^2 +14t -0.4 & \ge & 6 \\\\ -4.9t^2 +14t -0.4 & = & 6 \\ -4.9t^2 +14t -0.4 -6 & = & 0 \\ -4.9t^2 +14t -6.4 & = & 0 \\ t &=& \frac{-14\pm \sqrt{ 14^2-4\cdot (-4.9)\cdot (-6.4) }}{2\cdot(-4.9)} \\ t &=& \frac{-14\pm \sqrt{ 196-125.44 }}{-9.8} \\ t &=& \frac{-14\pm \sqrt{ 70.56 }}{-9.8} \\ t &=& \frac{-14\pm 8.4 }{-9.8} \\ t &=& \frac{14\pm 8.4 }{9.8} \\\\ \Delta t &=& t_2-t_1 \\ \Delta t &=& \frac{14 + 8.4 }{9.8}-\left(\frac{14- 8.4 }{9.8} \right) \\ \Delta t &=& \frac{14 + 8.4 -14+8.4}{9.8} \\ \Delta t &=& \frac{2\cdot 8.4}{9.8} \\ \Delta t &=& \frac{8.4}{4.9} \\ \Delta t &=& \frac{84}{49} \\ \Delta t &=& \frac{7\cdot 12}{7\cdot 7} \\ \mathbf{ \Delta t } & \mathbf{=} & \mathbf{\frac{ 12}{ 7}} \\ \hline \end{array}$$

The cannonball is $$\frac{12}{7}$$ seconds above a height of 6 meters

heureka  Feb 21, 2017

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