Question: Question 22
The vertical position $y$ (in feet) of a rock $t$ seconds after it was dropped from a cliff is given by the formula $y = -16t^2 - 10t + 450$. The base of the cliff corresponds to $y = 0$
After how many seconds will the rock hit the ground at the base of the cliff?
1. **State the problem:** We need to find the time $t$ when the rock hits the ground, which means when $y = 0$.
2. **Write the equation:** Given the formula for vertical position:
$$y = -16t^2 - 10t + 450$$
We set $y = 0$ to find when the rock hits the ground:
$$0 = -16t^2 - 10t + 450$$
3. **Solve the quadratic equation:**
$$-16t^2 - 10t + 450 = 0$$
Multiply both sides by $-1$ to simplify:
$$16t^2 + 10t - 450 = 0$$
4. **Use the quadratic formula:** For $at^2 + bt + c = 0$, the solutions are:
$$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
Here, $a = 16$, $b = 10$, and $c = -450$.
5. **Calculate the discriminant:**
$$\Delta = b^2 - 4ac = 10^2 - 4 \times 16 \times (-450) = 100 + 28800 = 28900$$
6. **Calculate the square root of the discriminant:**
$$\sqrt{28900} = 170$$
7. **Find the two possible values for $t$:**
$$t = \frac{-10 \pm 170}{2 \times 16} = \frac{-10 \pm 170}{32}$$
Calculate each:
- $$t_1 = \frac{-10 + 170}{32} = \frac{160}{32} = 5$$
- $$t_2 = \frac{-10 - 170}{32} = \frac{-180}{32} = -\frac{45}{8} = -5.625$$
8. **Interpret the results:**
Time cannot be negative, so we discard $t = -5.625$ seconds.
Therefore, the rock hits the ground after **$5$ seconds**.
**Final answer:** $5$ seconds