- The goats must clear a 4-meter wide crevasse to land at the same height on the cliff on the opposite side. - The goats jump at an angle of $$ 30^{\circ} $$ to the horizontal. - All goats have the same mass and height. That means that the quantities calculated are the same for all goats. Answer the following questions 1. What minimum launch speed must the goats have to ensure a safe landing? 2. What maximum height do the goats reach? 3. How long are the goats in the air? Use the following equations to solve the questions: Question 1 Equation: $$ R=\frac{r_{0}^{2} \sin \left(2 . \theta_{0}\right)}{9} $$ Question 2 Equation: $$ \Delta x=V_{0} x t $$ Question 3 Equation: $$ \Delta y=\operatorname{Voy} t+\frac{1}{2} g t^{2} $$

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**1. Minimum Launch Speed**

We are told that a goat must clear a crevasse of width $$ R=4 $$ meters, and the jump is made at an angle of $$\theta_0=30^\circ$$. The equation provided is

$$
R=\frac{V_0^2 \sin\left(2\theta_0\right)}{9}.
$$

Substitute $$\theta_0=30^\circ$$. Note that

$$
\sin(2\theta_0)=\sin(60^\circ)=\frac{\sqrt{3}}{2}.
$$

Hence the equation becomes

$$
4=\frac{V_0^2 \, (\sqrt{3}/2)}{9}.
$$

Solve for $$V_0^2$$ by multiplying both sides by 9:

$$
4\times9=\frac{V_0^2 \, (\sqrt{3}/2)}{1}\quad\Longrightarrow\quad 36=\frac{V_0^2 \sqrt{3}}{2}.
$$

Multiplying both sides by 2 gives

$$
72=V_0^2\sqrt{3}.
$$

Then

$$
V_0^2=\frac{72}{\sqrt{3}}.
$$

Rationalize the denominator by multiplying numerator and denominator by $$\sqrt{3}$$:

$$
V_0^2=\frac{72\sqrt{3}}{3}=24\sqrt{3}.
$$

Taking the square root we obtain

$$
V_0=\sqrt{24\sqrt{3}}.
$$

A numerical estimate is as follows. Since $$\sqrt{3}\approx1.732$$, then

$$
24\sqrt{3}\approx24\times1.732\approx41.57,
$$
$$
V_0\approx\sqrt{41.57}\approx6.45\text{ m/s}.
$$

---

**2. Maximum Height**

The maximum height reached in a projectile motion (when landing at the same height as launch) is given by

$$
h_{\text{max}}=\frac{\left(V_0\sin\theta_0\right)^2}{2g}.
$$

Here we use $$g=9\text{ m/s}^2$$ (as indicated by the use of 9 in the range equation) and $$\theta_0=30^\circ$$. The vertical component of the launch speed is

$$
V_{0y}=V_0\sin 30^\circ=\frac{V_0}{2}.
$$

Thus,

$$
h_{\text{max}}=\frac{\left(V_0/2\right)^2}{2\times9}=\frac{V_0^2}{4\times18}=\frac{V_0^2}{72}.
$$

Substitute the value $$V_0^2=24\sqrt{3}$$:

$$
h_{\text{max}}=\frac{24\sqrt{3}}{72}=\frac{\sqrt{3}}{3}.
$$

Numerically,

$$
h_{\text{max}}\approx\frac{1.732}{3}\approx0.577\text{ m}.
$$

---

**3. Time in the Air**

For a projectile landing at the same height as its launch point, the total time of flight is given by

$$
t=\frac{2V_0\sin\theta_0}{g}.
$$

Again, since $$V_0\sin30^\circ= \frac{V_0}{2}$$, we have

$$
t=\frac{2\,(V_0/2)}{9}=\frac{V_0}{9}.
$$

Using the computed value $$V_0\approx6.45\text{ m/s}$$:

$$
t\approx\frac{6.45}{9}\approx0.717\text{ s}.
$$

---

**Summary of Answers:**

1. Minimum launch speed:  
   $$
   V_0 = \sqrt{24\sqrt{3}} \approx 6.45\,\text{m/s}
   $$

2. Maximum height reached:  
   $$
   h_{\text{max}} = \frac{\sqrt{3}}{3} \approx 0.577\,\text{m}
   $$

3. Total time in the air:  
   $$
   t \approx 0.72\,\text{s}
   $$
**Answers:** 1. **Minimum Launch Speed:** The goats must launch at approximately **6.45 m/s** to safely clear the crevasse. 2. **Maximum Height Reached:** The goats reach a maximum height of about **0.577 meters**. 3. **Time in the Air:** The goats are airborne for approximately **0.72 seconds**.

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