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

t is between 0.47 seconds and 4.02 seconds

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 <0`

Solve t using quadratic formula:

`t = (-b \±√(b^2 - 4ac))/(2a)`

Where

`a = 4.9`

`b = -22`

`c = 9.25`

So, we have:

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

`t = (22 \±√(484 - 181.3))/(9.8)`

`t = (22 \±√(302.7))/(9.8)`

`t = (22 \±17.40)/(9.8)`

Split the equation

`t = (22 + 17.40)/(9.8)` or
`t = (22 - 17.40)/(9.8)`

`t = (39.4)/(9.8)` or
`t = (4.6)/(9.8)`

`t = (39.4)/(9.8)` or
`t = (4.6)/(9.8)`

`t = 4.02` or
`t = 0.47`

Hence; the range is

`0.47 < t <4.02`