Question

a cannon ball is launched into the air with an upward velocity of 116 feet per second, from a 6 - foot tall cannon. the height h of the cannon ball t seconds after launch can be found using the equation h(t)=-16t² + 116t + 6. at what time(s) after launch will the cannon ball be 108 feet high? round answer(s) to the nearest tenth if necessary. how long will it take to hit the ground?

Explanation:

Step1: Set up the equation for height 108 feet

Set $h(t)=108$, so $- 16t^{2}+116t + 6=108$. Rearrange to get $-16t^{2}+116t - 102 = 0$. Divide by -2: $8t^{2}-58t + 51=0$.

Step2: Use the quadratic formula

The quadratic formula for $ax^{2}+bx + c = 0$ is $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here, $a = 8$, $b=-58$, $c = 51$. First, calculate the discriminant $\Delta=b^{2}-4ac=(-58)^{2}-4\times8\times51=3364 - 1632=1732$. Then $t=\frac{58\pm\sqrt{1732}}{16}=\frac{58\pm41.62}{16}$.
$t_1=\frac{58 + 41.62}{16}=\frac{99.62}{16}\approx6.2$ and $t_2=\frac{58 - 41.62}{16}=\frac{16.38}{16}\approx1.0$.

Step3: Set up the equation for hitting the ground

When the cannon - ball hits the ground, $h(t)=0$. So $-16t^{2}+116t + 6 = 0$. Divide by -2: $8t^{2}-58t - 3=0$.

Step4: Use the quadratic formula again

For $8t^{2}-58t - 3=0$, $a = 8$, $b=-58$, $c=-3$. Calculate the discriminant $\Delta=b^{2}-4ac=(-58)^{2}-4\times8\times(-3)=3364+96 = 3460$. Then $t=\frac{58\pm\sqrt{3460}}{16}=\frac{58\pm58.82}{16}$. We take the positive root since time cannot be negative. $t=\frac{58 + 58.82}{16}=\frac{116.82}{16}\approx7.3$.

Answer:

When the cannon - ball is 108 feet high: $t\approx1.0$ seconds, $t\approx6.2$ seconds
When the cannon - ball hits the ground: $t\approx7.3$ seconds