Question

A ball is thrown into the air. The path it takes is modeled by the equation: -3t+24t = h, where t is the time in seconds and h is the height of the ball above the ground, measured in feet. Write an inequality to model when the height of the ball is at least 36 feet above the ground. For how long is the ball at or above 36 feet?​

Answer 1

Given:

The given equation is:

$-3t^2+24t=h$

Where, t is the time in seconds and h is the height of the ball above the ground, measured in feet.

To find:

The inequality to model when the height of the ball is at least 36 feet above the ground. Then find time taken by ball to reach at or above 36 feet.

Solution:

We have,

$-3t^2+24t=h$

The height of the ball is at least 36 feet above the ground. It means $h\geq 36$.

$-3t^2+24t\geq 36$

$-3t^2+24t-36\geq 0$

$-3(t^2-8t+12)\geq 0$

Splitting the middle term, we get

$-3(t^2-6t-2t+12)\geq 0$

$-3(t(t-6)-2(t-6))\geq 0$

$-3(t-2)(t-6)\geq 0$

The critical points are:

$-3(t-2)(t-6)=0$

$t=2,6$

These two points divide the number line in 3 intervals $(-\infty,2),(2,6),(6,\infty)$.

Intervals      Check point           $-3(t-2)(t-6)\geq 0$           Result

$(-\infty,2)$               0                       $(-)(-)(-)=(-)$         False

$(2,6)$                    4                       $(-)(+)(-)=+>0$            True

$(6,\infty)$                  8                        $(-)(+)(+)=(-)$        False

The inequality is true for (2,6) and the sign of inequality is $\geq$. So, the ball is above 36 feet between 2 to 6 seconds.

$6-2=4$

Therefore, the required inequality is $-3t^2+24t\geq 36$ and the ball is 36 feet above for 4 seconds.