Question

ASAP!! Please help me. I will not accept nonsense answers, but will mark as BRAINLIEST if you answer is correctly with solutions.

Answer 1

Answer:

$t = 2.86 \: s$

The ball remains in the air for 2.86 seconds.

Step-by-step explanation:

A baseball is hit into the air, and its height above the ground is described by the function

$H(t) = -16t^2 + 45t + 2$

Where H(t) represents the height of the ball t seconds after it is hit.

The ball remains in the air before it hits the ground.

When the ball hits the ground,

$H(t) = 0$

So,

$0 = -16t^2 + 45t + 2$

The equation may be solved using the quadratic formula.

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

Where

$a = -16 \\\\b = 45 \\\\c = 2$

So,

$t=\frac{-(45)\pm\sqrt{(45)^2-4(-16)(2)}}{2(-16)} \\\\t=\frac{-45\pm\sqrt{(2025 - (-128)}}{-32} \\\\t=\frac{-45\pm\sqrt{(2153}}{-32} \\\\t=\frac{-45\pm 46.4004}{-32} \\\\t=\frac{-45 + 46.4004}{-32} \: and \: t=\frac{-45 - 46.4004}{-32}\\\\t= -0.04 \: and \: t = 2.86$

Since the time cannot be negative, discard the negative value and accept the positive value

$t = 2.86 \: s$

Therefore, the ball remains in the air for 2.86 seconds.