+0

# Help with algebra

0
72
1

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
+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