Type the correct answer in the box. Use numerals instead of words.

A ball is kicked with an initial height of 0.75 meters and initial upward velocity of 22 meters/second. This inequality represents the time, [tex]t[/tex], in seconds, when the ball's height is greater than 10 meters.

[tex]\[ -4.9 t^2 + 22 t + 0.75 \ \textgreater \ 10 \][/tex]

The ball's height is greater than 10 meters when [tex]t[/tex] is approximately between [tex]\square[/tex] and [tex]\square[/tex] seconds.



Answer :

To solve the inequality:

[tex]\[ -4.9 t^2 + 22 t + 0.75 > 10 \][/tex]

we need to follow these steps:

1. Simplify the inequality:

Subtract 10 from both sides to set the inequality to zero:

[tex]\[ -4.9 t^2 + 22 t + 0.75 - 10 > 0 \][/tex]

Simplify further:

[tex]\[ -4.9 t^2 + 22 t - 9.25 > 0 \][/tex]

2. Find the roots of the corresponding quadratic equation:

Solve the equation:

[tex]\[ -4.9 t^2 + 22 t - 9.25 = 0 \][/tex]

We can use the quadratic formula for this:

[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]

where [tex]\( a = -4.9 \)[/tex], [tex]\( b = 22 \)[/tex], and [tex]\( c = -9.25 \)[/tex].

Calculate the discriminant:

[tex]\[ b^2 - 4ac = 22^2 - 4(-4.9)(-9.25) = 484 - 4(4.9)(9.25) \][/tex]

Simplify the discriminant:

[tex]\[ b^2 - 4ac = 484 - 181.3 = 302.7 \][/tex]

Take the square root of the discriminant:

[tex]\[ \sqrt{302.7} \approx 17.4 \][/tex]

Now, substitute back into the quadratic formula:

[tex]\[ t = \frac{-22 \pm 17.4}{-9.8} \][/tex]

Find the two solutions:

[tex]\[ t_1 = \frac{22 - 17.4}{9.8} \approx 0.469 \][/tex]

[tex]\[ t_2 = \frac{22 + 17.4}{9.8} \approx 4.02 \][/tex]

3. Determine the intervals where the inequality holds:

The parabola [tex]\( -4.9 t^2 + 22 t - 9.25 \)[/tex] opens downward (because the coefficient of [tex]\( t^2 \)[/tex] is negative). Therefore, the height will be greater than 10 meters between the two roots.

Thus, the ball’s height is greater than 10 meters when [tex]\( t \)[/tex] is approximately between [tex]\( 0.469 \)[/tex] and [tex]\( 4.02 \)[/tex] seconds.

The correct answer, when the ball's height is greater than 10 meters, is:

[tex]\[ \boxed{0.469 \text{ and } 4.02} \][/tex]