Answer
Solution : Between 0 and 3 seconds
Work Step by Step
Given: The position function $s (t) = $-16$t^{2}+ v_{0} t+ s_{0}$ $> s_{0}.
$Here, $v_{0}$ = 48 feet per second$ ; s_{0} = $160 feet$.
$In order to find the period of time during which the ball’s height exceeds that of the rooftop, let us solve the function
$s (t) = $-16$t^{2} $+ $v_{0} t $+ $s_{0}$
As $s (t) $> $s_{0}$
On substituting the given data, we have
$s (t) = $-16$t^{2}+ v_{0} t+ s_{0}$ $> s_{0}$
Here are the steps required for Solving Polynomial Inequalities:
1. One side must be zero and the other side can have only one fraction, so we simplify the fractions if there is more than one fraction.
Let us subtract 160 from both sides.
Thus,
$-16 t^{2}$$+ $48$t$ $ > 0$
2. Solve the equation $f (x)$ $=$ $0$. The real solutions are the boundary points.
$-16 t^{2}$$+$ $48 t$ $=$ $0$
$-8$ $t$ $($$2$$t$-6) = 0
$-8$ $t$ $=$ $0$ or $2$$t$ $-$ $6$ $=$ $0$
This implies
$t$ $=$ $0$ or $t$ $=$ $3$
3. Locate the boundary points on a number line found in Step 2 to divide the number line into intervals.
The boundary points divide the number line into three intervals:
For our purposes the mathematical model is useful only from $t$ $=$ $0$ until the diver hits the ground.
Let us determine the time when the diver hits the ground.
$s$ ($t$) $=$ $0$
$-16$ $t^{2}$$+$ $v_{0}$ $t$ $+$ $s_{0}$ $=$ $0$
$-16$ $t^{2}$$+$ 48 $t$ $+$ $160$ $=$ $0$
This implies
$t$ $≈$ $5$
As $t$ $≥$ $0$ only $t$ $=$ $5$ fits. Therefore, we use the intervals
(0, 3), (3, 5).
4. Test each interval’s sign of $f (t)$ with a test value.
Interval Test value Substitute into
$ f(t)$ $=$ $- 16 t^{2}$ $+$ $48$ $t$
$(0, 3)$ $1$ $f (t) > 0$
Positive
$(3, 5)$ 4
Negative $f (t) < 0$
5. Write the solution set, selecting the interval or intervals that satisfy the given inequality. Solve, $f (t) > 0$, where $f (t) = - 16 t^{2} + 48 t$ . Based on our work done in Step 4, we see that $f (t) > 0$ for all $t$ in
$(0 , 3)$.
Conclusion: This means that the diver’s height exceeds that of the rooftop between 0 and 3 second.
