Calculating Impulsive Force on Baseball Player's Hand

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
8 replies · 6K views
jaredogden
Messages
77
Reaction score
0

Homework Statement



A baseball player catching a ball can soften the impact by pulling
his hand back. Assuming that a 5-oz ball reaches his glove at
90 mi/h and that the player pulls his hand back during the impact
at an average speed of 30 ft/s over a distance of 6 in., bringing the
ball to a stop, determine the average impulsive force exerted on
the player’s hand.

Homework Equations



mv1 + Imp1→2 = mv2
v2 = v02 + 2a(x - x0
x = x0 + v0t + 1/2at2

The Attempt at a Solution



(o ft/s)2 = (132 ft/s)2 + 2a(6/12 ft)
a = -17421 ft/s2

(6/12 ft) = 0 ft + 132 ft/st + 1/2(-17424 ft/s2)t2
t = 0.007575s

0.3125 lbs (132 ft/s) + F(0.007575s) = 0.3125 lbs (0 ft/s)
F = 54446 lbs

The answer is wayyyy less as it should be..
76.9 lbs

I just can't figure out how to get there..
 
Physics news on Phys.org
oh sorry

0 = -8712 ft/s2t2 + (135 ft/s)t + 0.5 ft

t = [-(135 ft/s) + sqrt((135 ft/s)2) - (4*(-8712 ft/s2)*(0.5 ft))]/(2*(-8712 ft/s2))

t = -0.003088 s


t = [-(135 ft/s) - sqrt((135 ft/s)2) - (4*(-8712 ft/s2)*(0.5 ft))]/(2*(-8712 ft/s2))

t = 0.0185841 s

mv1 + Imp1→2 = mv2
0.3125 lbs (132 ft/s) + F(0.0185841 s) = 0.3125 lbs (0 ft/s)
F = 2219.639
F = 2220 lbs

I'm still getting the wrong answer.
 
Oh I put = + 0.5 ft.. in my calculator I did -0.5 ft

the answer for t is still the same.

I'm obviously missing something in my approach..

jaredogden said:
The answer is wayyyy less as it should be..
76.9 lbs
 
Ah of course... I'm so stupid. Yeah there is, mass in English units is in slugs not lbs. I should be dividing .3125 lbs by 32.2 ft/s2

I'm closer... but still off by about 6.4 lbs.. I must be forgetting something..

mv1 + Imp1→2 = mv2
(0.3125 lbs/(32.2 ft/s2)) (132 ft/s) + F(0.0185841 s) = 0.3125 lbs (0 ft/s)
F = 70.499 lbs
F = 70.5 lbs

The answer is 76.9

This is frustrating, I feel like an idiot.
 
I just realized in the second time I did the quadratic formula I was using 135 ft/s instead of 132 ft/s.

Also used that for the linear momentum.. I'm redoing everything right now.