A baseball catcher puts on an exhibition by catching a 0.10-kg ball dropped from a helicopter at a height of 81 m.What is the speed of the ball just before it hits the catcher's glove 1 m above the ground? (g = 9.8 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 formula v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity (0 m/s), a is the acceleration due to gravity (9.8 m/s^2), and s is the distance traveled (80 m). Plugging in these values, we get v^2 = 0^2 + 2(9.8)(80), which gives v = 39.6 m/s.

A Baseball Catcher Puts on an Exhibition by Catching a 0.10-Kg