+0  
 
0
72
1
avatar

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 8.7 meters?

 Jun 15, 2021
 #1
avatar
+1

For quadratic \(ax^2+bx-c\)\(a < 0\). Its graph is a negative parabola. Let's calculate when the cannonball will fly exactly 8.7 meters. Substitute t into \(h(t) = -4.9t^2+14t-0.4\) which \(h(t)\) is 8.7 meters.  

Assume \(x = t\)

\(-4.9x^2+14x-0.4=8.7\)

\(-4.9x^2+14x-9.1=0\)

\(4.9x^2-14x+9.1=0\)

Substitute into the quadratic formula\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\).

We get\(x = {-(-14) \pm \sqrt{(-14)^2-4*4.9*9.1} \over 2*4.9}\)

\(x = {14 \pm \sqrt{196-178.36} \over 9.8}\)

\(x = {14 \pm \sqrt{17.64} \over 9.8}\)

\(x = {14 \pm 4.2 \over 9.8}\)

\({x}_{1} = \frac{13}{7}\) or \({x}_{2} = 1\)

the region is \(\left |{x}_{2} - {x}_{1} \right |\), which equals \(\frac{6}{7}\)

 Jun 15, 2021

15 Online Users

avatar
avatar