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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,t, in seconds, when the ball's height is greater than 10 meters. -4.9t^2+22t+0.75>10.
the ball's height is greater than 10 meters when t is approximately between blank and blank seconds​

Answer :

MrRoyal

Answer:

t is between 0.47 seconds and 4.02 seconds

Step-by-step explanation:

Given

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

Required

Determine the values of t

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

Subtract 10 from both sides

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

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

Multiply through y -1

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

Solve t using quadratic formula:

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

Where

[tex]a = 4.9[/tex]

[tex]b = -22[/tex]

[tex]c = 9.25[/tex]

So, we have:

[tex]t = \frac{-(-22) \±\sqrt{(-22)^2 - 4*4.9*9.25}}{2 * 4.9}[/tex]

[tex]t = \frac{22 \±\sqrt{484 - 181.3}}{9.8}[/tex]

[tex]t = \frac{22 \±\sqrt{302.7}}{9.8}[/tex]

[tex]t = \frac{22 \±17.40}{9.8}[/tex]

Split the equation

[tex]t = \frac{22 + 17.40}{9.8}[/tex] or [tex]t = \frac{22 - 17.40}{9.8}[/tex]

[tex]t = \frac{39.4}{9.8}[/tex] or [tex]t = \frac{4.6}{9.8}[/tex]

[tex]t = \frac{39.4}{9.8}[/tex] or [tex]t = \frac{4.6}{9.8}[/tex]

[tex]t = 4.02[/tex] or [tex]t = 0.47[/tex]

Hence; the range is

[tex]0.47 < t <4.02[/tex]

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