A pitcher throws a 0.139-kg baseball, and it approaches thebat at a speed of 47.

Question

A pitcher throws a 0.139-kg baseball, and it approaches thebat at a speed of 47.3 m/s. The bat does 84.8 J of work on the ballin hitting it. Ignoring air resistance, determine the speed of theball after the ball leaves the bat and is 28.7 m above the point ofimpact. Given : .             Mass of the ball (m) = 0.139 kg .             Initial velocity (v1)   = 47.3 m/s .             Work done (W)   = 84.8 J .              Distance moved (d)   = 28.7 m . Hence    Force is :   F  = W / d   = 84.8 J / 28.7   = 2.955 N . From Work - Energy theorem : .              F * d   = ( 1/2 ) m   [(v2)2   -   (v1)2 ] .                         = ( 0.5 * 0.139 ) [ (v2)2  -   (47.3)2 ] .           (v2)2       =   [ (F * d ) / ( 0.5 * 0.139) ] +   (47.3)2 .            v2          =  ---------- m /s Okay so I followed this and I got 58.8 as the finalvelocity, but its marked wrong - ' can anyone explain how else tosolve this question?? - or what am i doing wrong? A pitcher throws a 0.139-kg baseball, and it approaches thebat at a speed of 47.3 m/s. The bat does 84.8 J of work on the ballin hitting it. Ignoring air resistance, determine the speed of theball after the ball leaves the bat and is 28.7 m above the point ofimpact. Given : .             Mass of the ball (m) = 0.139 kg .             Initial velocity (v1)   = 47.3 m/s .             Work done (W)   = 84.8 J .              Distance moved (d)   = 28.7 m . Hence    Force is :   F  = W / d   = 84.8 J / 28.7   = 2.955 N . From Work - Energy theorem : .              F * d   = ( 1/2 ) m   [(v2)2   -   (v1)2 ] .                         = ( 0.5 * 0.139 ) [ (v2)2  -   (47.3)2 ] .           (v2)2       =   [ (F * d ) / ( 0.5 * 0.139) ] +   (47.3)2 .            v2          =  ---------- m /s Okay so I followed this and I got 58.8 as the finalvelocity, but its marked wrong - ' can anyone explain how else tosolve this question?? - or what am i doing wrong?

Explanation / Answer

1) Find initial kinetic energy. KE =.5mv2.   2) Add 85 J to that kinetic energy (because thebat did work on it) 3) Find the potential energy at height (h) = 28.7m.   PE = mgh. 4) Subtract this Potential energy from the kinetic energyfound in step 2. 5) Find the velocity by using your kinetic energy. KE =.5mv2   and you solve for v. Doing all of that, I got 53.8 .

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