Physics
posted by Omari on .
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)
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?

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