A baseball catcher puts on an exhibition by catching a 0.150 kg ball dropped from a helicopter at a height of 100 meters above the catcher. If the catcher "gives" with the ball for a distance of 0.750 meters while catching it, what average force is exerted on the mitt by the ball? (g = 9.80 m/s^2)

Question

a. 78N
b. 119N
c. 197N
d. 392N

The equation I have for average force is "Favg=delta (m*v) / delta time" but I don't think I can apply it directly to this situation.

How do you solve this problem?

Answer 1

You could use your equation, but would need one more step to be able to use the "give" distance X instead of the "delta time" of deceleration.
The average velocity of the ball while being decelerated is v/2. The "give" distance is thus
X = (v/2)*(delta t)

You will also need
v = sqrt(2gh) = 44.3 m/s
(neglecting air friction)

Favg = m*v/(2x/v) = (1/2)mv^2/X
= 197 N