an arrow is fired into the air with an initial upward velocity of 96 feet per second from the top of a…
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
Answer
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