4) A rocket is launched from a 5-foot tall platform at 40 -feet per second When does the rocket hit the ground? $$ h(t)=-16 t^{2}+v_{0} t+h_{0} $$

Question

UpStudy AI · Accepted Answer

**Step 1.** Write the height equation and set it equal to zero to find when the rocket hits the ground:

$$
-16t^2 + 40t + 5 = 0
$$

**Step 2.** Multiply the entire equation by $$-1$$ to simplify:

$$
16t^2 - 40t - 5 = 0
$$

**Step 3.** Identify the coefficients for the quadratic equation $$at^2 + bt + c = 0$$:

$$
a = 16, \quad b = -40, \quad c = -5
$$

**Step 4.** Use the quadratic formula:

$$
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
$$

Substitute the coefficients into the formula:

$$
t = \frac{-(-40) \pm \sqrt{(-40)^2 - 4(16)(-5)}}{2(16)}
$$

$$
t = \frac{40 \pm \sqrt{1600 + 320}}{32}
$$

$$
t = \frac{40 \pm \sqrt{1920}}{32}
$$

**Step 5.** Simplify the square root. Notice that:

$$
\sqrt{1920} = \sqrt{64 \times 30} = 8\sqrt{30}
$$

Now substitute back:

$$
t = \frac{40 \pm 8\sqrt{30}}{32}
$$

**Step 6.** Factor out 8 from the numerator:

$$
t = \frac{8(5 \pm \sqrt{30})}{32} = \frac{5 \pm \sqrt{30}}{4}
$$

**Step 7.** Choose the positive solution (since time cannot be negative):

$$
t = \frac{5 + \sqrt{30}}{4}
$$

Thus, the rocket hits the ground at:

$$
\boxed{\frac{5 + \sqrt{30}}{4} \text{ seconds}}
$$
The rocket hits the ground after $$ \frac{5 + \sqrt{30}}{4} $$ seconds.

A rocket is launched from a 5-foot tall
platform at 40 -feet per second When does
the rocket hit the ground?