A baseball catcher puts on an exhibition by catching a 0.200-kg ball dropped from a helicopter at a height of 150 m above the catcher.If the catcher "gives" with the ball for a distance of 0.755 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 of the ball when dropped can be found using:
PE = mgh
where m = 0.200 kg, g = 9.80 m/s^2, and h = 150 m
PE = (0.200 kg)(9.80 m/s^2)(150 m) = 294 J

When the ball is caught, it undergoes a change in momentum. We can find the final velocity of the ball just before it is caught using:
PE = (1/2)mv^2
where v is the velocity just before the ball is caught.

Rearranging this equation and solving for v, we get:
v = sqrt((2PE)/m)
v = sqrt((2*294 J)/(0.200 kg)) = 60.9 m/s

The time it takes the ball to fall from 150 m can be found using:
h = (1/2)gt^2
where h = 150 m and g = 9.80 m/s^2
Solving for t, we get:
t = sqrt((2h)/g) = sqrt((2*150 m)/(9.80 m/s^2)) = 5.03 s

The average force exerted on the mitt can then be found using:
F = (mv)/t
where m = 0.200 kg, v = 60.9 m/s, and t = 5.03 s

F = (0.200 kg * 60.9 m/s) / 5.03 s = 2.42 N

However, we need to account for the "give" of 0.755 m. This means that the mitt moves 0.755 m while the ball is being caught, increasing the time it takes for the mitt to come to a stop. We can use the equation of motion to find this time:
d = (1/2)at^2
where d = 0.755 m, a = F/m, and t is the time it takes for the mitt to come to a stop.

Rearranging this equation and solving for t, we get:
t = sqrt((2d)/(F/m))
t = sqrt((2*0.755 m)/((2.42 N)/(0.200 kg))) = 0.837 s

The average force exerted on the mitt including the "give" is then:
F_avg = (mv)/(t + 0.837 s)
F_avg = (0.200 kg * 60.9 m/s)/(5.03 s + 0.837 s) = 389 N

Therefore, the answer is C) 389 N.

A Baseball Catcher Puts on an Exhibition by Catching a 0.200-Kg