Question

The height h (in feet) of an underhand volleyball serve can be modeled by h=−16t2+30t+4, where t is the time in seconds. After how many seconds is the ball 16 feet above the ground?

Answer 1

Answer:

$t\approx 0.58, t\approx 1.30$

Step-by-step explanation:

We have been given that the height h (in feet) of an underhand volleyball serve can be modeled by $h=-16t^2+30t+4$, where t is the time in seconds.

To find the time when ball will be 16 feet above the ground, we will substitute $h=16$ in our given equation an solve for t as:

$16=-16t^2+30t+4$

$-16t^2+30t+4-16=0$

$-16t^2+30t-12=0$

Using quadratic formula, we will get:

$t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$t=\frac{-30\pm\sqrt{30^2-4(-16)(-12)}}{2(-16)}$

$t=\frac{-30\pm\sqrt{900-768}}{-32}$

$t=\frac{-30\pm\sqrt{132}}{-32}$

$t=\frac{-30\pm 11.48912529}{-32}$

$t=\frac{-30+11.48912529}{-32}, t=\frac{-30-11.48912529}{-32}$

$t=\frac{-18.51087471}{-32}, t=\frac{-41.48912529}{-32}$

$t=0.5784648, t=1.296535$

$t\approx 0.58, t\approx 1.30$

Therefore, the ball will be 16 feet above the ground after 0.58 seconds or 1.30 seconds.

Answer 2

After 0.57 second or 1.29 seconds height of the ball will be 16 feet.

To understand more, check below explanation.

Height of ball:

The height of ball is modeled by function,

                     $h=-16t^{2} +30t+4$

Where t is time.

We have to find time for which height of ball is 16 feet.

                        $-16t^{2} +30t+4=16\\\\-16t^{2} +30t-12=0\\\\t=\frac{-30\pm\sqrt{900-192} }{-32} \\\\t=0.57,t=1.29$

Therefore, after 0.57 second or 1.29 seconds height of the ball will be 16 feet.

Learn more about the function here:

https://brainly.com/question/25638609