Let's break down the problem step by step. ### **Problem Statement** - The maximum horizontal distance (range) a boy can throw a ball is \( R = 40 \) m. - He can throw with the same initial speed at all angles. - **Question:** How high does he throw the ball when he throws it straight upward? --- ## **Step 1: Find the Initial Speed (\(v_\))** The formula for the range of a projectile (thrown at an angle \(\theta\)) is: \[ R = \frac{v_^2 \sin 2\theta}{g} \] The maximum range is achieved at \(\theta = 45^\circ\), so \(\sin 2\theta = \sin 90^\circ = 1\): \[ R_{\text{max}} = \frac{v_^2}{g} \] Given \(R_{\text{max}} = 40\) m, and \(g = 9.8\, \text{m/s}^2\): \[ 40 = \frac{v_^2}{9.8} \] \[ v_^2 = 40 \times 9.8 = 392 \] \[ v_ = \sqrt{392} \approx 19.8\, \text{m/s} \] --- ## **Step 2: Find the Maximum Height When Thrown Straight Up** When thrown straight up, all the initial speed is in the vertical direction. The maximum height \(H\) is given by: \[ H = \frac{v_^2}{2g} \] Plug in the values: \[ H = \frac{(19.8)^2}{2 \times 9.8} \] \[ H = \frac{392}{19.6} \] \[ H = 20 \, \text{m} \] --- ## **Final Answer** \[ \boxed{20\, \text{m}} \] --- ### **Summary** - **Initial speed found from range:** \(v_ \approx 19.8\, \text{m/s}\) - **Maximum height when thrown straight up:** \( \boxed{20\, \text{m}} \)