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?
+26367 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
+128474 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 ]
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.
+26367 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
