Respuesta :
Answer:
a. 1 second
b. 9 seconds
Explanation:
Part A.
To find the time when the ball reaches the height of 128 ft, we put h(t) = 128 into our equation and get
[tex]128=-16t^2+144t[/tex]We rearrange the above equation to write it in a more familiar form by subtracting 128 from both sides
[tex]-16t^2+144t-128=0[/tex]The above equation is quadratic, and therefore, we can solve for t using the quadratic formula.
The quadratic formula gives
[tex]t=\frac{-144\pm\sqrt[]{114^2-4(-16)(-128)}}{2(-16)}[/tex]which upon simplification gives us the following solutions.
[tex]\begin{gathered} t=1 \\ t=8 \end{gathered}[/tex]Hence, 1 and 8 seconds after the throw, the ball is at a height of 128 ft.
Since we want the earliest time, from the two values of t we choose t = 1 as our relevant answer.
Part B.
When the ball hits the ground, its height above the ground is zero. Therefore, we put h= 0 into our formula and get
[tex]0=-16t^2+144t[/tex]To solve for t, we first add 16t^2 to both sides and get
[tex]16t^2=144t[/tex]dividing both sides by 16t gives
[tex]\frac{16t^2}{16t}=\frac{144t}{16t}[/tex][tex]\boxed{t=9.^{}}[/tex]Hence, the ball hits the ground 9 seconds after the launch.
