A pitcher throws a 0.145-kg 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!
The work of the hit equals the change of kinetic energy of the ball before and after the hit. So, you can calculate the speed of the ball after the hit by using that fact. Further on, since total mechanical energy is conserved, you can use the fact that the sum of potential and kinetic energy of the ball right after the hit equals the sum of potential and kinetic energy which the ball has at the height of 28.5 m to retrieve the speed of the ball at the mentioned height.
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??
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!
Exactly! You would set that total energy which you just found equal to KEafter + PEafter. (Assuming you set PE = 0 when the ball is at the bat.) Then solve for v^2 like you did before.