# Problem Overview - Tim stands on the ground and throws a ball straight up. - Dr. J is on a balcony **3.7 m above Tim**. - Dr. J sees the ball pass her **0.4 s after the throw**, while the ball is still rising. - Tim catches the ball when it comes back down. - **Find:** 1. **Maximum height** the ball reaches above Tim. 2. **Total time** the ball spends in the air. --- # Variables and Constants - Let \( y = 0 \) be Tim's hand (launch point). - \( y_1 = 3.7\,\mathrm{m} \): Dr. J's balcony. - \( t_1 = 0.4\,\mathrm{s} \): Time when ball passes Dr. J. - \( g = 9.8\,\mathrm{m/s}^2 \): Acceleration due to gravity. - \( v_0 \): Ball's initial velocity (upwards). - \( t_{\text{max}} \): Time to reach max height. - \( t_{\text{total}} \): Total time in air. - \( h_{\text{max}} \): Max height above Tim. --- # Step 1: Find Initial Velocity \( v_0 \) Use the kinematic equation for vertical motion: \[ y = v_0 t - \frac{1}{2} g t^2 \] When ball passes Dr. J (\( y = 3.7\,\mathrm{m} \), \( t = 0.4\,\mathrm{s} \)): \[ 3.7 = v_0 \cdot 0.4 - \frac{1}{2} (9.8) (0.4)^2 \] \[ 3.7 = 0.4 v_0 - 0.5 \times 9.8 \times 0.16 \] \[ 3.7 = 0.4 v_0 - 0.784 \] \[ 0.4 v_0 = 3.7 + 0.784 = 4.484 \] \[ v_0 = \frac{4.484}{0.4} = 11.21\,\mathrm{m/s} \] --- # Step 2: Maximum Height (\( h_{\text{max}} \)) At the maximum height, \( v = 0 \): \[ v = v_0 - g t_{\text{max}} = 0 \implies t_{\text{max}} = \frac{v_0}{g} \] \[ t_{\text{max}} = \frac{11.21}{9.8} = 1.144\,\mathrm{s} \] Height at \( t_{\text{max}} \): \[ h_{\text{max}} = v_0 t_{\text{max}} - \frac{1}{2} g t_{\text{max}}^2 \] \[ = 11.21 \times 1.144 - 0.5 \times 9.8 \times (1.144)^2 \] \[ = 12.832 - 0.5 \times 9.8 \times 1.309 \] \[ = 12.832 - 6.414 \] \[ = 6.42\,\mathrm{m} \] --- # Step 3: Total Time in Air (\( t_{\text{total}} \)) When the ball returns to Tim's hand (\( y = 0 \)): \[ 0 = v_0 t_{\text{total}} - \frac{1}{2} g t_{\text{total}}^2 \] \[ 0 = t_{\text{total}} (v_0 - \frac{1}{2} g t_{\text{total}}) \] \[ \implies t_{\text{total}} = 0 \quad \text{or} \quad v_0 - \frac{1}{2} g t_{\text{total}} = 0 \] But actually, set up as: \[ 0 = v_0 t_{\text{total}} - \frac{1}{2} g t_{\text{total}}^2 \] \[ v_0 t_{\text{total}} = \frac{1}{2} g t_{\text{total}}^2 \] \[ v_0 = \frac{1}{2} g t_{\text{total}} \] \[ t_{\text{total}} = \frac{2 v_0}{g} \] \[ t_{\text{total}} = \frac{2 \times 11.21}{9.8} = 2.29\,\mathrm{s} \] --- # Final Answers ## 1. Maximum Height **The ball reaches a maximum height of \( \boxed{6.4\,\mathrm{m}} \) above Tim's hand.** ## 2. Total Time in Air **The ball is in the air for \( \boxed{2.29\,\mathrm{s}} \).**