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.

Answer 1

Complete 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.

-4.9t² + 22t + 0.75 > 10

The ball's height is greater than 10 meters when t is approximately between __ and __ seconds

Answer:

The ball's height is greater than 10 meters when t is approximately between 0.47 and 4.02 seconds

Step-by-step explanation:

Given

$-4.9t\² + 22t + 0.75 > 10$

Required

Solve the inequality

$-4.9t\² + 22t + 0.75 > 10$

Subtract 10 from both sides

$-4.9t\² + 22t + 0.75 - 10> 10 - 10$

$-4.9t\² + 22t + -9.25> 0$

Solve using quadratic formula

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

In this case;

a = -4.9; b = 22; c = -9.25

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

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

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

$t = \frac{22 \± \sqrt{302.7}}{9.8}$

$t = \frac{22 \± 17.398}{9.8}$

Split

$t = \frac{22 + 17.398}{9.8}$    or      $t = \frac{22 - 17.398}{9.8}$

$t = \frac{39.398}{9.8}$   or     $t = \frac{4.602}{9.8}$

$t = 4.02$ or  $t = 0.47$

Convert to inequality;

$0.47 < t < 4.02$

Hence;

The ball's height is greater than 10 meters when t is approximately between 0.47 and 4.02 seconds