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.

To find the speed of the girl and the box just after she grabs the box at height \( h = 2.00 \) meters, we can use the principle of conservation of momentum and the kinematic equations. First, calculate the speed of the girl just before she grabs the box. We use the kinematic equation: \[ v^2 = v_i^2 + 2gh \] Given: - \( v_i = 8.00 \) m/s (initial velocity) - \( g = -9.80 \) m/s² (negative because it acts downward) - \( h = 2.00 \) m Calculate the final velocity \( v \): \[ v^2 = (8.00)^2 + 2(-9.80)(-2.00) \] \[ v^2 = 64 + 39.2 = 103.2 \] \[ v = \sqrt{103.2} \approx 10.16 \, \text{m/s} \] Now, using conservation of momentum: Let \( v_f \) be the final velocity after she grabs the box. The initial momentum before she grabs the box is: \[ p_i = m_1 \cdot v + m_2 \cdot 0 = 60.0 \cdot 10.16 + 15.0 \cdot 0 \] \[ p_i = 609.6 \, \text{kg m/s} \] After grabbing the box, the total mass is \( m_{total} = m_1 + m_2 = 60.0 + 15.0 = 75.0 \) kg. The final momentum \( p_f \) is given by: \[ p_f = m_{total} \cdot v_f \] Setting the initial momentum equal to the final momentum: \[ 609.6 = 75.0 \cdot v_f \] Solve for \( v_f \): \[ v_f = \frac{609.6}{75.0} \approx 8.13 \, \text{m/s} \] So, the speed of the girl and the box just after she grabs the box is approximately \( 8.13 \) meters per second.