\begin{tabular}{l} A girl of mass \( m_{1}=60.0 \) kilograms springs from a trampoline with an initial \\ upward velocity of \( v_{i}=8.00 \) meters per second. At height \( h=2.00 \) meters \\ above the trampoline, the girl grabs a box of mass \( m_{2}=15.0 \) kilograms. ( \\ Eigure 1) \\ For this problem, use \( g=9.80 \) meters per second per second for the \\ magnitude of the acceleration due to gravily. \\ What is the maximum height \( h_{\text {max }} \) that the girl (with box) reaches? Measure \( h_{\text {max }} \) with respect to the top of the trampoline. \\ Express your answer numerically in meters. \\ Part \\ \hline\end{tabular}

Below is the step‐by‐step solution in markdown format. --- ### Step 1. Determine the Girl’s Speed at \( h = 2.00 \) m The girl starts from the trampoline with an upward speed of \[ v_i = 8.00\ \text{m/s}. \] At a height of \( 2.00 \) meters above the trampoline, we can find her speed \( v \) using energy conservation or the kinematics relation \[ v^2 = v_i^2 - 2g h. \] Plug in the given values: \[ v^2 = (8.00)^2 - 2(9.80)(2.00) = 64.0 - 39.2 = 24.8. \] Thus, the speed is \[ v = \sqrt{24.8} \approx 4.98\ \text{m/s}. \] --- ### Step 2. Apply Conservation of Momentum When the Girl Grabs the Box At the height of \( 2.00 \) m, the girl (mass \( m_1 = 60.0 \) kg) grabs a box (mass \( m_2 = 15.0 \) kg). Immediately after grabbing the box, by conservation of momentum the total momentum is conserved. Before grabbing the box, the momentum is \[ p_{\text{initial}} = m_1 \, v = 60.0 \times 4.98 \approx 298.8\ \text{kg·m/s}. \] After the collision the total mass is \[ m_{\text{total}} = m_1 + m_2 = 60.0 + 15.0 = 75.0\ \text{kg}. \] Let \( v_{\text{after}} \) be the new upward speed. Conservation of momentum gives \[ m_1 \, v = (m_1 + m_2) v_{\text{after}}, \] \[ v_{\text{after}} = \frac{60.0 \times 4.98}{75.0} \approx \frac{298.8}{75.0} \approx 3.98\ \text{m/s}. \] For simplicity, we can round this to \[ v_{\text{after}} \approx 4.00\ \text{m/s}. \] --- ### Step 3. Calculate the Additional Height Gained After the Collision After the collision, the combined mass rises with an initial speed of \( 4.00 \) m/s. The additional height \( \Delta h \) that the combined system will rise can be found using energy considerations (or the kinematic equation with final speed zero): \[ \Delta h = \frac{v_{\text{after}}^2}{2g}. \] Substitute the known values: \[ \Delta h = \frac{(4.00)^2}{2(9.80)} = \frac{16.0}{19.6} \approx 0.8163\ \text{m}. \] --- ### Step 4. Determine the Maximum Height with Respect to the Trampoline The maximum height reached above the trampoline is the sum of the height at collision and the additional height: \[ h_{\text{max}} = 2.00\ \text{m} + 0.8163\ \text{m} \approx 2.82\ \text{m}. \] --- ### Final Answer The maximum height \( h_{\text{max}} \) that the girl (with the box) reaches, measured with respect to the top of the trampoline, is \[ \boxed{2.82\ \text{meters}}. \]