Let's break the problem down step by step: ## **Given:** - Distance to wall, \( x = 100 \) m - Height of wall, \( y = 15 \) m - Initial speed, \( v_0 = 80 \) m/s - Acceleration due to gravity, \( g = 9.8 \) m/s² ## **Step 1: Projectile Motion Equations** The position of the projectile at time \( t \): - **Horizontal (x):** \( x = v_0 \cos\theta \cdot t \) - **Vertical (y):** \( y = v_0 \sin\theta \cdot t - \frac{1}{2}gt^2 \) ## **Step 2: Time to Reach the Wall** Solve for \( t \) when \( x = 100 \) m: \[ t = \frac{x}{v_0 \cos\theta} \] ## **Step 3: Height of Projectile at the Wall** Plug \( t \) into the \( y \) equation: \[ y = v_0 \sin\theta \cdot t - \frac{1}{2}g t^2 \] \[ y = v_0 \sin\theta \left(\frac{x}{v_0 \cos\theta}\right) - \frac{1}{2}g \left(\frac{x}{v_0 \cos\theta}\right)^2 \] \[ y = x \tan\theta - \frac{g x^2}{2 v_0^2 \cos^2\theta} \] Set \( y = 15 \) m: \[ 15 = 100 \tan\theta - \frac{9.8 \times 100^2}{2 \times 80^2 \cos^2\theta} \] \[ 15 = 100 \tan\theta - \frac{98000}{12800 \cos^2\theta} \] \[ 15 = 100 \tan\theta - \frac{7.65625}{\cos^2\theta} \] Rearrange: \[ 100 \tan\theta - \frac{7.65625}{\cos^2\theta} - 15 = 0 \] ## **Step 4: Solve for \( \theta \)** Let \( \tan\theta = t \), and \( \cos^2\theta = 1/(1+t^2) \): \[ 100 t - 7.65625 (1 + t^2) - 15 = 0 \] \[ 100 t - 7.65625 - 7.65625 t^2 - 15 = 0 \] \[ 100 t - 7.65625 t^2 - 22.65625 = 0 \] \[ 7.65625 t^2 - 100 t + 22.65625 = 0 \] ## **Step 5: Quadratic Formula** \[ t = \frac{100 \pm \sqrt{100^2 - 4 \times 7.65625 \times 22.65625}}{2 \times 7.65625} \] \[ t = \frac{100 \pm \sqrt{10000 - 693.326}}{15.3125} \] \[ t = \frac{100 \pm \sqrt{9306.674}}{15.3125} \] \[ t = \frac{100 \pm 96.464}{15.3125} \] ### **Two solutions:** 1. \( t_1 = \frac{100 + 96.464}{15.3125} \approx \frac{196.464}{15.3125} \approx 12.8 \) 2. \( t_2 = \frac{100 - 96.464}{15.3125} \approx \frac{3.536}{15.3125} \approx 0.231 \) ## **Step 6: Find the Angles** \[ \theta_1 = \arctan(12.8) \approx 85.5^\circ \] \[ \theta_2 = \arctan(0.231) \approx 13.0^\circ \] ### **Interval Notation:** \[ \boxed{\theta \in [13.0, 85.5]} \] --- ### **Final Answer:** \[ \boxed{\theta \in [13.0, 85.5]} \] (Set your catapult to any angle between **13.0** and **85.5** degrees for the rocks to clear the wall.)