Answer:
[tex]t = 2.86 \: s[/tex]
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
[tex]H(t) = -16t^2 + 45t + 2[/tex]
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,
[tex]H(t) = 0[/tex]
So,
[tex]0 = -16t^2 + 45t + 2[/tex]
The equation may be solved using the quadratic formula.
[tex]$t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$[/tex]
Where
[tex]a = -16 \\\\b = 45 \\\\c = 2 \\\\[/tex]
So,
[tex]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 \\\\[/tex]
Since the time cannot be negative, discard the negative value and accept the positive value
[tex]t = 2.86 \: s[/tex]
Therefore, the ball remains in the air for 2.86 seconds.
