Sure! Let’s break down the problem step by step. ### **Understanding the Problem** - **Given:** An oscillating garden sprinkler discharges water with an **initial velocity** \( v_0 = 8 \) m/s. - **Goal:** Find the **distance \( d \)** (how far the water reaches) and the **angle \( \alpha \)** (the angle at which the water should be launched). #### **Given Data:** - Initial speed, \( v_0 = 8 \) m/s. - Horizontal distance to the start of the garden, \( 1.5 \) m. - We want the farthest point \( B \) (total distance \( d \) from point \( A \)), and the angle \( \alpha \). --- ### **Step 1: Understand Projectile Motion** When the water is thrown at an angle \( \alpha \), it follows a **parabolic path**. The horizontal distance it covers is called the **range**. #### **Projectile Range Formula** The horizontal range \( R \) (in the absence of air resistance) is: \[ R = \frac{v_0^2 \sin(2\alpha)}{g} \] where: - \( v_0 \) is the initial velocity, - \( \alpha \) is the launch angle, - \( g \) is the acceleration due to gravity (\( 9.81 \) m/s²). --- ### **Step 2: Calculate the Farthest Distance \( d \)** The maximum range is when \( \sin(2\alpha) \) is maximum, which happens at \( 2\alpha = 90^\circ \) (i.e., \( \alpha = 45^\circ \)). So, \[ d_{\text{max}} = \frac{v_0^2}{g} \] Plug in the numbers: \[ d_{\text{max}} = \frac{8^2}{9.81} = \frac{64}{9.81} \approx 6.525 \text{ m} \] --- ### **Step 3: Find the Angle \( \alpha \)** For **maximum range**, the angle is: \[ \alpha = 45^\circ \] --- ### **Step 4: Adjust for Initial Horizontal Distance** If you want the distance from the start of the garden (point \( A \)) to the farthest point \( B \) *after* 1.5 m, you **add** 1.5 m to your answer only if the origin is *before* the garden. But based on the diagram, it looks like \( d \) is measured from the sprinkler, so you don't need to add 1.5 m unless specifically asked for the distance from the garden's edge. --- ### **Final Answers** - **The distance \( d \) to the farthest point \( B \) is:** \[ \boxed{6.525\ \text{m}} \] - **The angle \( \alpha \) is:** \[ \boxed{45^\circ} \] --- #### **Summary Table:** | Quantity | Formula | Value | |------------------|----------------------------------|----------| | Maximum range \( d \) | \( \frac{v_0^2}{g} \) | 6.525 m | | Angle \( \alpha \) | \( 45^\circ \) | 45° | --- If you need to consider the 1.5 m (for example, the garden starts 1.5 m from the sprinkler), then subtract 1.5 m from the total calculated range. Let me know if the garden starts at a different point, or if you have more details!