A baseball catcher puts on an exhibition by catching a 0.175-kg ball dropped from a helicopter at a height of 89.0 m.What is the speed of the ball just before it hits the catcher's glove 1.00 m above the ground? (g = 9.80 m/s² and ignore air resistance)

Question

Quizplus · Accepted Answer

The correct answer is A because the speed of the ball just before it hits the catcher's glove can be calculated using the conservation of energy principle. The potential energy of the ball at the height of 89.0 m is converted into kinetic energy just before it hits the catcher's glove. The kinetic energy of the ball is given by:$$KE = 1/2 mv^2$$where m is the mass of the ball and v is its speed. The potential energy of the ball at the height of 89.0 m is given by:$$PE = mgh$$where g is the acceleration due to gravity and h is the height. The conservation of energy principle states that the total energy of the system (PE + KE) is constant. Therefore, we have:$$PE = KE$$$$mgh = 1/2 mv^2$$Solving for v, we get:$$v = sqrt(2gh)$$Substituting the given values, we get:$$v = sqrt(2 * 9.80 m/s^2 * 89.0 m) = 41.5 m/s$$

A Baseball Catcher Puts on an Exhibition by Catching a 0.175-Kg