Question

midterm review score: 89.99/100 answered: 17/18 question 18 a cannon ball is launched into the air with an upward velocity of 128 feet per second, from a cannon on a high mountain 1344 feet from the ground. the height h of the cannon ball after t seconds can be found using the equation h(t)=-16t² + 128t + 1344. approximately how long will it take for the cannon ball to be 1600 feet high? round answers to the nearest tenth if necessary. select an answer how long will the cannon ball take to hit the ground? select an answer

Explanation:

Step1: Set up the first - part equation

Set $h(t)=1600$, so $-16t^{2}+128t + 1344=1600$. Rearrange it to the standard quadratic form $ax^{2}+bx + c = 0$. We get $-16t^{2}+128t+1344 - 1600=0$, which simplifies to $-16t^{2}+128t - 256 = 0$. Divide through by - 16 to get $t^{2}-8t + 16=0$.

Step2: Solve the quadratic equation

For the quadratic equation $t^{2}-8t + 16 = 0$, where $a = 1$, $b=-8$, $c = 16$. Use the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. First, calculate the discriminant $\Delta=b^{2}-4ac=(-8)^{2}-4\times1\times16=64 - 64=0$. Then $t=\frac{8\pm\sqrt{0}}{2}=\frac{8}{2}=4$ seconds.

Step3: Set up the second - part equation

When the cannon - ball hits the ground, $h(t)=0$. So $-16t^{2}+128t + 1344=0$. Divide through by - 16 to get $t^{2}-8t - 84=0$.

Step4: Solve the second quadratic equation

For the quadratic equation $t^{2}-8t - 84=0$, where $a = 1$, $b=-8$, $c=-84$. Calculate the discriminant $\Delta=b^{2}-4ac=(-8)^{2}-4\times1\times(-84)=64 + 336 = 400$. Then $t=\frac{8\pm\sqrt{400}}{2}=\frac{8\pm20}{2}$. We have two solutions: $t_1=\frac{8 + 20}{2}=14$ and $t_2=\frac{8-20}{2}=-6$. Since time cannot be negative, we discard $t_2$.

Answer:

The cannon - ball is 1600 feet high at $t = 4$ seconds. The cannon - ball hits the ground at $t = 14$ seconds.