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. ( Figure 1) For this problem, use \( g=9.80 \) meters per second per second for the magnitude of the acceleration due to gravity. Express your answer numerically in meters per second.

Below is a step‐by‐step solution in markdown format. --- ### Step 1. Determine the Girl’s Speed at \( h = 2.00 \) m The girl is launched with an initial upward velocity of \[ v_i = 8.00 \text{ m/s}. \] As she rises, gravity decelerates her motion. Using energy conservation (or kinematics), we have: \[ v^2 = v_i^2 - 2gh, \] where - \( g = 9.80 \text{ m/s}^2 \), - \( h = 2.00 \text{ m} \). Plug in the values: \[ v^2 = (8.00)^2 - 2(9.80)(2.00) = 64.0 - 39.2 = 24.8. \] Taking the square root gives the upward speed at \( h = 2.00 \) m: \[ v = \sqrt{24.8} \approx 4.98 \text{ m/s}. \] --- ### Step 2. Apply Conservation of Momentum During the Collision At \( h = 2.00 \) m, the girl (mass \( m_1 = 60.0 \) kg) grabs a stationary box (mass \( m_2 = 15.0 \) kg). Since the collision is perfectly inelastic (they stick together), linear momentum is conserved. Before the collision, the total momentum is: \[ p_{\text{initial}} = m_1 v \quad \text{(since the box is at rest)}. \] After the collision, the total mass is \[ m_1 + m_2, \] and they move together with velocity \( v_f \). Setting the momenta equal: \[ m_1 v = (m_1 + m_2) v_f. \] Solving for \( v_f \): \[ v_f = \frac{m_1 v}{m_1 + m_2}. \] Substitute the known values: \[ v_f = \frac{60.0 \text{ kg} \times 4.98 \text{ m/s}}{60.0 \text{ kg} + 15.0 \text{ kg}} = \frac{298.8 \text{ kg·m/s}}{75.0 \text{ kg}} \approx 3.98 \text{ m/s}. \] --- ### Final Answer The combined system (girl plus box) moves upward with a velocity of approximately: \[ \boxed{3.98 \text{ m/s}}. \]