A baseball catcher puts on an exhibition by catching a 0.170-kg ball dropped from a helicopter at a height of 110 m above the catcher.If the catcher "gives" with the ball for a distance of 0.695 m while catching it,what average force is exerted on the mitt by the ball? (g = 9.80 m/s²)

Question

Quizplus · Accepted Answer

First, we need to find the velocity of the ball just before it hits the mitt. Using the kinematic equation:

v^2 = u^2 + 2as

where v = final velocity (0 m/s, since the ball comes to rest), u = initial velocity (unknown), a = acceleration due to gravity (-9.80 m/s^2), and s = distance fallen (110 m).

Rearranging, we get:

u = sqrt(v^2 - 2as)

u = sqrt(-2(-9.80)(110)) = 47.0 m/s

Now, we can use the work-energy theorem to find the average force exerted on the mitt:

W = Fd = KEf - KEi

where W = work done by the mitt on the ball (negative since the force and displacement are in opposite directions), F = average force exerted on the mitt, d = distance the mitt "gives" (0.695 m), KEf = final kinetic energy of the ball (0 J), and KEi = initial kinetic energy of the ball (1/2 mv^2).

Rearranging, we get:

F = (KEi - KEf) / d = (1/2 mv^2) / d

F = (1/2)(0.170 kg)(47.0 m/s)^2 / 0.695 m = 264 N

Therefore, the answer is C) 264 N.

A Baseball Catcher Puts on an Exhibition by Catching a 0.170-Kg