Question

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 1

Answer:

t is between 0.47 seconds and 4.02 seconds

Step-by-step explanation:

Given

$Inequality: -4.9t^2+22t+0.75>10$

Required

Determine the values of t

$-4.9t^2+22t+0.75>10$

Subtract 10 from both sides

$-4.9t^2+22t+0.75 - 10 >10 - 10$

$-4.9t^2+22t-9.25 >0$

Multiply through y -1

$4.9t^2-22t+9.25$

Solve t using quadratic formula:

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

Where

$a = 4.9$

$b = -22$

$c = 9.25$

So, we have:

$t = \frac{-(-22) \±\sqrt{(-22)^2 - 4*4.9*9.25}}{2 * 4.9}$

$t = \frac{22 \±\sqrt{484 - 181.3}}{9.8}$

$t = \frac{22 \±\sqrt{302.7}}{9.8}$

$t = \frac{22 \±17.40}{9.8}$

Split the equation

$t = \frac{22 + 17.40}{9.8}$ or $t = \frac{22 - 17.40}{9.8}$

$t = \frac{39.4}{9.8}$ or $t = \frac{4.6}{9.8}$

$t = \frac{39.4}{9.8}$ or $t = \frac{4.6}{9.8}$

$t = 4.02$ or $t = 0.47$

Hence; the range is

$0.47 < t$