Let's break down the problem and solve it step by step: --- ## **Given Data** - **Initial height (\( y_A \))**: \( 2 \) m - **Final height (\( y_B \))**: \( 3 \) m - **Horizontal distance**: \( 10 \) m - **Launch angle (\( \theta_A \))**: \( 30^\circ \) - **Gravity (\( g \))**: \( 9.81 \) m/s² --- ## **Step 1: Break Initial Velocity into Components** Let \( v_A \) be the initial velocity. - Horizontal: \( v_{Ax} = v_A \cos 30^\circ \) - Vertical: \( v_{Ay} = v_A \sin 30^\circ \) --- ## **Step 2: Time to Reach the Basket (Horizontal Motion)** The horizontal motion is **uniform** (no acceleration): \[ x = v_{Ax} t \implies 10 = (v_A \cos 30^\circ) t \implies t = \frac{10}{v_A \cos 30^\circ} \] --- ## **Step 3: Vertical Motion Equation** Use the second equation of motion: \[ y_B = y_A + v_{Ay} t - \frac{1}{2} g t^2 \] Plug in values: \[ 3 = 2 + (v_A \sin 30^\circ) t - \frac{1}{2} (9.81) t^2 \] \[ 1 = (v_A \cdot .5) t - 4.905 t^2 \] --- ## **Step 4: Substitute \( t \) from Step 2** From Step 2: \[ t = \frac{10}{v_A \cos 30^\circ} \] \[ \cos 30^\circ = .866 \] \[ t = \frac{10}{v_A \cdot .866} \] Substitute into the vertical equation: \[ 1 = (v_A \cdot .5) \left( \frac{10}{v_A \cdot .866} \right) - 4.905 \left( \frac{10}{v_A \cdot .866} \right)^2 \] \[ 1 = \frac{5}{.866} - 4.905 \cdot \frac{100}{v_A^2 \cdot .866^2} \] \[ 1 = 5.773 - \frac{490.5}{v_A^2 \cdot .75} \] \[ 1 = 5.773 - \frac{654}{v_A^2} \] \[ \frac{654}{v_A^2} = 5.773 - 1 = 4.773 \] \[ v_A^2 = \frac{654}{4.773} \] \[ v_A^2 = 137.03 \] \[ v_A = \sqrt{137.03} = \boxed{11.7 \text{ m/s}} \] --- ## **Step 5: Find the Velocity at the Basket (\( v_B \))** ### **Horizontal Component at B** Remains the same: \[ v_{Bx} = v_A \cos 30^\circ = 11.7 \times .866 = 10.13 \text{ m/s} \] ### **Vertical Component at B** \[ v_{By} = v_{Ay} - g t \] First, find \( t \): \[ t = \frac{10}{v_A \cdot .866} = \frac{10}{11.7 \times .866} = \frac{10}{10.13} = .987 \text{ s} \] Now, \[ v_{Ay} = v_A \sin 30^\circ = 11.7 \times .5 = 5.85 \text{ m/s} \] \[ v_{By} = 5.85 - 9.81 \times .987 = 5.85 - 9.69 = -3.84 \text{ m/s} \] ### **Magnitude of \( v_B \):** \[ v_B = \sqrt{v_{Bx}^2 + v_{By}^2} \] \[ v_B = \sqrt{10.13^2 + (-3.84)^2} = \sqrt{102.6 + 14.75} = \sqrt{117.35} = \boxed{10.8 \text{ m/s}} \] --- ## **Step 6: Direction of \( v_B \) Below the Horizontal** \[ \theta = \tan^{-1} \left( \frac{|v_{By}|}{v_{Bx}} \right) = \tan^{-1} \left( \frac{3.84}{10.13} \right) = \tan^{-1}(.379) = 20.7^\circ \] --- ## **Final Answers** \[ \boxed{ \begin{aligned} \text{Initial velocity, } v_A &= 11.7 \text{ m/s} \\ \text{Velocity at basket, } v_B &= 10.8 \text{ m/s} \\ \text{Angle below horizontal, } \theta &= 20.7^\circ \end{aligned} } \] Let me know if you want a visual breakdown or further clarification!