- 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: \[ \Delta y=\operatorname{Voy} t+\frac{1}{2} g t^{2} \]

To find the minimum launch speed required for the goats to clear the 4-meter-wide crevasse while jumping at a \(30^{\circ}\) angle, we can substitute into the range equation. Rearranging the range equation for \(R\): \[ R = \frac{V_0^2 \sin(2 \theta_0)}{g} \] This gives us: \[ V_0^2 = \frac{R \cdot g}{\sin(2 \theta_0)} \] Now let's evaluate this. Using \(R = 4 \text{ m}\), \(g \approx 9.81 \text{ m/s}^2\), and \(\theta_0 = 30^{\circ}\), we have \(\sin(2 \cdot 30^{\circ}) = \sin(60^{\circ}) = \frac{\sqrt{3}}{2}\): \[ V_0^2 = \frac{4 \cdot 9.81}{\frac{\sqrt{3}}{2}} \] Calculating this gives us \(V_0^2 \approx \frac{39.24}{0.866} \approx 45.3\) which means: \[ V_0 \approx \sqrt{45.3} \approx 6.73 \text{ m/s} \] Now on to the maximum height: 1. The vertical component of the velocity is: \[ V_{0y} = V_0 \sin(\theta) = 6.73 \cdot \frac{1}{2} = 3.365 \text{ m/s} \] 2. The maximum height \(\Delta y\) can be found using the kinematic equation where the final velocity is \(0\): \[ 0 = V_{0y}^2 + 2g\Delta y \] Rearranging this gives us: \[ \Delta y = \frac{-V_{0y}^2}{2g} = \frac{-(3.365)^2}{2 \cdot (-9.81)} \approx \frac{11.32}{19.62} \approx 0.577 \text{ m} \] To find out how long the goats are in the air, we need to calculate the time it takes to reach maximum height and then double it (as the ascent and descent times are equal). The time to reach maximum height can be calculated from: \[ t = \frac{V_{0y}}{g} = \frac{3.365}{9.81} \approx 0.342 \text{ s} \] Thus, the total time in the air is: \[ T = 2t = 2 \cdot 0.342 \approx 0.684 \text{ s} \] So, the answers are: 1. Minimum launch speed \( V_0 \approx 6.73 \text{ m/s} \) 2. Maximum height \( \Delta y \approx 0.577 \text{ m} \) 3. Time in the air \( T \approx 0.684 \text{ s} \)