(30) A man drops a \( 2.7 \cdot \mathrm{~kg} \) bag out a window 11 meters above ground to his friend, 1.7 meters above ground. Determine the velocity of the bag just before it reaches his friend's hands. Notebook

To find the velocity of the bag just before it reaches his friend's hands, we can use the principle of conservation of mechanical energy. The potential energy lost by the bag will convert into kinetic energy. The initial height of the bag is \( h_i = 11 \, \text{m} \) and the height of the friend is \( h_f = 1.7 \, \text{m} \). The height it falls is: \[ h = h_i - h_f = 11 \, \text{m} - 1.7 \, \text{m} = 9.3 \, \text{m} \] The potential energy at the initial height is given by: \[ PE = mgh_i \] where \( g \) (acceleration due to gravity) is approximately \( 9.81 \, \text{m/s}^2 \). The change in potential energy when moving from height \( h_i \) to height \( h_f \) is converted into kinetic energy (KE): \[ KE = \frac{1}{2}mv^2 \] Setting the change in potential energy equal to kinetic energy, we have: \[ mgh = \frac{1}{2}mv^2 \] Cancelling \( m \) from both sides (since \( m \) is not zero) gives: \[ gh = \frac{1}{2}v^2 \] Solving for \( v \), we get: \[ v = \sqrt{2gh} \] Substituting in the values: \[ v = \sqrt{2 \cdot 9.81 \, \text{m/s}^2 \cdot 9.3 \, \text{m}} = \sqrt{182.73} \approx 13.53 \, \text{m/s} \] Thus, the velocity of the bag just before it reaches his friend's hands is approximately \( 13.53 \, \text{m/s} \).