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Kinetic energy of a baseball 
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#1
Oct406, 12:57 PM

P: 157

A pitcher throws a 0.145kg baseball, and it approaches the bat at a speed of 53.6 m/s. The bat does Wnc = 77.4 J of work on the ball in hitting it. Ignoring the air resistance, determine the speed of the ball after the ball leaves the bat and is 28.5 m above the point of impact.
The only thing I can think of here is that the x displacement is zero since the ball is hit straight up. this means the Cos(theta) is equal to 1. Other than that I am not sure where to begin. I've done similar problems to this, but I'm not so sure how to incorporate 77.4 J into any of my equations...Any ideas? Thanks! 


#2
Oct406, 01:06 PM

HW Helper
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#3
Oct906, 11:26 AM

P: 157

My attempt at solving the problem:
KEbefore = .5*.145*53.6^2= 208.2896 J PEbefore = 0 208.2896= .5*.145*v^2 + 40.5398 + 77.4 =.0725*v^2=90.3498 v^2=1246.2 v=35.30 m/s This is wrong but I'm unsure why?? 


#4
Oct906, 11:45 AM

Mentor
P: 41,316

Kinetic energy of a baseball
Answer this: What's the total energy of the ball immediately after getting hit by the bat?



#5
Oct906, 01:09 PM

P: 157

Total energy after hitting the ball would be: 208.2896 J + 77.4 J (KEbefore + Energy from the bat) Then wouldn't you just take KEbefore + PEbefore and set it equal to KEafter + PEafter, then solve for v^2 like I did above? Sorry, I'm basically learning this from what I've read in my book; my teacher is being replaced on Wednesday for receiving such a low grade on student evaluations and a number of complaints by the student body as a whole (suprise suprise). Thanks for the help! 


#6
Oct906, 01:40 PM

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#7
Oct906, 02:25 PM

P: 157

v=58.20, thanks Doc Al!!



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