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 heigh…

Question

asked Dec 17, 2021 222k views

2 Answers

Wibosco · Answer 1 · 2021-12-21T11:02:41+0000

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

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

$t = (-b \± √(b^2 - 4ac))/(2a)$

In this case;

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

$t = (-22 \± √(22^2 - 4 * (-4.9) * (-9.25)))/(2 * -4.9)$

$t = (-22 \± √(484 - 181.3))/(-9.8)$

$t = (22 \± √(484 - 181.3))/(9.8)$

$t = (22 \± √(302.7))/(9.8)$

$t = (22 \± 17.398)/(9.8)$

Split

t = (22 + 17.398)/(9.8) or
t = (22 - 17.398)/(9.8)

t = (39.398)/(9.8) or
t = (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