A baseball catcher puts on an exhibition by catching a \( 0.15-\mathrm{kg} \) ball dropped from a helicopter at a height of 101 m . What is the speed of the ball just before it hits the catcher's glove 1.0 m above the ground? \( \left(g=9.8 \mathrm{~m} / \mathrm{s}^{2}\right. \) and ignore air resistance) 1) \( 44 \mathrm{~m} / \mathrm{s} \) 2) \( 38 \mathrm{~m} / \mathrm{s} \) 3) \( 31 \mathrm{~m} / \mathrm{s} \) 4) \( 22 \mathrm{~m} / \mathrm{s} \) 5) \( 15 \mathrm{~m} / \mathrm{s} \)

To find the speed of the ball just before it hits the catcher's glove, we can use the conservation of energy principle. The potential energy (PE) at the height of 101 m is converted into kinetic energy (KE) as the ball falls. The height from which the ball is dropped is 101 m, but since it is caught 1 m above the ground, the ball falls from 100 m (101 m - 1 m = 100 m). The potential energy at 100 m is given by: \[ PE = mgh \] Where: - \( m = 0.15 \, \text{kg} \) - \( g = 9.8 \, \text{m/s}^2 \) - \( h = 100 \, \text{m} \) Calculating the potential energy: \[ PE = 0.15 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 100 \, \text{m} = 147 \, \text{J} \] This potential energy will be converted entirely into kinetic energy just before the ball hits the catcher's glove: \[ KE = \frac{1}{2} mv^2 \] Setting PE equal to KE: \[ 147 \, \text{J} = \frac{1}{2} \times 0.15 \, \text{kg} \times v^2 \] Solving for \( v \): \[ 147 = 0.075 v^2 \] \[ v^2 = \frac{147}{0.075} = 1960 \] \[ v = \sqrt{1960} \approx 44.4 \, \text{m/s} \] Thus, the speed of the ball just before it hits the catcher's glove is approximately \( 44 \, \text{m/s} \). Therefore, the correct answer is: 1) \( 44 \, \text{m/s} \).