- The goats must clear a 4-meter wide crevasse ta land at the same height on the cliff on the opposite side. - The goats fump 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 sir? Use the following equations to solve the questions: Question 1 Equation: $$ R=\frac{-2 \sin \left(2 \theta_{0}\right)}{9} $$ $$ \sqrt{\frac{4 \times 98}{\sin (60)}}=6.73 $$ Question 2 Equation: $$ \begin{array}{l} \Delta x=V_{0} x t \ \text { razontial displament IQuation } T=\frac{\Delta x}{V_{0} x} \frac{4 m}{-673 / m 5}-0.59 \end{array} $$ nilestion 3 Equation: $$ \text { Find } \Delta y \text { using This Equation (Max height } $$ $$ \Delta_{y}=V_{0 y}++\frac{1}{2} g t^{2} $$

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1. We start with the projectile‐motion range formula for a launch and landing at equal heights:
   
   $$
   R = \frac{v_{0}^{2} \sin(2\theta)}{g}
   $$
   
   Here, $$R=4\,\text{m}$$, $$\theta=30^\circ$$ (so $$\sin(2\theta)=\sin(60^\circ)$$), and $$g=9.8\,\text{m/s}^2$$. Solving for $$v_0$$:
   
   $$
   v_{0} = \sqrt{\frac{R\,g}{\sin(2\theta)}} = \sqrt{\frac{4 \times 9.8}{\sin(60^\circ)}}
   $$
   
   Since $$\sin(60^\circ) \approx 0.866$$, we have:
   
   $$
   v_{0} = \sqrt{\frac{39.2}{0.866}} \approx \sqrt{45.3} \approx 6.73\,\text{m/s}
   $$
   
2. The maximum height reached in projectile motion is found using the vertical component. The formula is:
   
   $$
   h_{\text{max}} = \frac{v_{0}^{2} \sin^{2}(\theta)}{2g}
   $$
   
   With $$\theta=30^\circ$$ ($$\sin(30^\circ)=0.5$$) and $$v_{0}\approx6.73\,\text{m/s}$$:
   
   $$
   h_{\text{max}} = \frac{(6.73)^2 \times (0.5)^2}{2 \times 9.8} = \frac{(45.3) \times 0.25}{19.6} \approx \frac{11.33}{19.6} \approx 0.578\,\text{m}
   $$
   
3. The total time of flight for a projectile launched and landing at the same height is given by:
   
   $$
   T = \frac{2\,v_{0} \sin(\theta)}{g}
   $$
   
   Substituting the values $$v_{0}\approx6.73\,\text{m/s}$$ and $$\sin(30^\circ)=0.5$$:
   
   $$
   T = \frac{2 \times 6.73 \times 0.5}{9.8} = \frac{6.73}{9.8} \approx 0.686\,\text{s}
   $$
   
Thus, the answers are:

1. The minimum launch speed is approximately $$6.73\,\text{m/s}$$.
2. The maximum height reached is approximately $$0.58\,\text{m}$$.
3. The time in the air is approximately $$0.69\,\text{s}$$.
1. The goats must launch at a minimum speed of about 6.73 m/s to safely land. 2. They reach a maximum height of approximately 0.58 meters. 3. They are in the air for about 0.69 seconds.

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