Calculating Impulsive Force on Baseball Player's Hand

  • #1
jaredogden
79
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..
 
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  • #2
jaredogden said:
(6/12 ft) = 0 ft + 132 ft/st + 1/2(-17424 ft/s2)t2
t = 0.007575s
This is a quadratic; not correctly solved.
 
  • #3
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.
 
  • #4
jaredogden said:
0 = -8712 ft/s2t2 + (135 ft/s)t + 0.5 ft
I believe one of these terms has the wrong sign.

I'm still getting the wrong answer.
It is always helpful, when you know the correct answer, to provide it so others know their working is right.
 
  • #5
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
 
  • #6
I've almost forgotten how to work in imperial measures.
Isn't there a factor of 32.2 because F is in poundals https://www.physicsforums.com/images/icons/icon5.gif
 
Last edited by a moderator:
  • #7
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.
 
  • #8
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.
 
  • #9
Yes, I noticed your 135 mistake. Still doesn't quite get me to 76 though. I used 0.0183 sec
 

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