Question

a cannon ball is launched into the air with an upward velocity of 64 feet per second, from a cannon on a high mountain 512 feet from the ground. the height h of the cannon ball after t seconds can be found using the equation h(t)=-16t² + 64t + 512. approximately how long will it take for the cannon ball to be 576 feet high? round answers to the nearest tenth if necessary. how long will the cannon ball take to hit the ground?

Explanation:

Step1: Set up the equation for the first - part

We want to find when $h(t)=576$. So we set $-16t^{2}+64t + 512=576$. Rearrange it to the standard quadratic form $ax^{2}+bx + c = 0$.
$-16t^{2}+64t+512 - 576=0$, which simplifies to $-16t^{2}+64t - 64 = 0$. Divide through by - 16 to get $t^{2}-4t + 4 = 0$.

Step2: Solve the quadratic equation

For a quadratic equation $at^{2}+bt + c = 0$ (here $a = 1$, $b=-4$, $c = 4$), we can use the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. But $t^{2}-4t + 4=(t - 2)^{2}=0$. So $t = 2$ seconds.

Step3: Set up the equation for the second - part

We want to find when $h(t)=0$. So we set $-16t^{2}+64t + 512=0$. Divide through by - 16 to get $t^{2}-4t - 32=0$.

Step4: Factor the quadratic equation

Factor $t^{2}-4t - 32=(t - 8)(t+4)=0$.

Step5: Find the values of t

Set each factor equal to zero: $t - 8=0$ gives $t = 8$ and $t+4=0$ gives $t=-4$. Since time cannot be negative in this context, $t = 8$ seconds.

Answer:

1. 2 seconds
2. 8 seconds