Question

an arrow is fired into the air with an initial upward velocity of 96 feet per second from the top of a building 256 feet high. the equation that gives the hight of the arrow at any time t is h = 256 + 96t - 16t². find the times at which the arrow will be 384 feet in the air? the arrow will be 384 feet in the air after and seconds. question help: video ebook

Explanation:

Step1: Set up the equation

Set $h = 384$ in the equation $h=256 + 96t-16t^{2}$. So we get $384=256 + 96t-16t^{2}$.

Step2: Rearrange the equation

Rearrange it to the standard - form of a quadratic equation $ax^{2}+bx + c = 0$.
$16t^{2}-96t + 128 = 0$. Divide through by 16 to simplify: $t^{2}-6t + 8 = 0$.

Step3: Factor the quadratic equation

Factor the left - hand side: $(t - 2)(t - 4)=0$.

Step4: Solve for t

Set each factor equal to zero:
If $t - 2=0$, then $t = 2$.
If $t - 4=0$, then $t = 4$.

Answer:

2, 4