Question

arthur drops a ball from a height of 81 feet above the ground. its height, h, is given by the equation $h = -16t^2 + 81$, where t is the time in seconds. for which interval of time is the height of the ball less than 17 feet?\
\\(\circ\\) $t > 2$\
\\(\circ\\) $0 < t < 2$\
\\(\circ\\) $t = 2$\
\\(\circ\\) $-2 < t < 2$

Explanation:

Step1: Set up inequality for height

$h < 17$ so substitute $h=-16t^2+81$:
$$-16t^2 + 81 < 17$$

Step2: Rearrange the inequality

Subtract 81 from both sides:
$$-16t^2 < 17 - 81$$
$$-16t^2 < -64$$

Step3: Isolate $t^2$ (reverse inequality)

Divide both sides by -16:
$$t^2 > \frac{-64}{-16}$$
$$t^2 > 4$$

Step4: Solve for t, discard negative time

Square root both sides; time $t>0$:
$$t > 2$$

Answer:

t>2