A baseball catcher puts on an exhibition by catching a 0.150-kg ball dropped from a helicopter at a height of 100 m above the catcher.If the catcher "gives" with the ball for a distance of 0.500 m while catching it,what average force is exerted on the mitt by the ball? (g = 9.80 m/s²)

Question

Quizplus · Accepted Answer

The potential energy at 100 m is converted to kinetic energy just before the catch. Using energy conservation: mgh = 0.5mv^2. Solving for v gives v = sqrt(2gh). The work done by the force is equal to the change in kinetic energy: F*d = 0.5mv^2. Solving for F gives F = (0.5mv^2)/d. Substituting the values, F = (0.5 * 0.150 * (sqrt(2 * 9.8 * 100))^2) / 0.5 = 295 N.

A Baseball Catcher Puts on an Exhibition by Catching a 0.150-Kg