Find the speed of the ball after the ball leaves the bat

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To find the speed of the baseball after it leaves the bat, the initial speed before impact, the work done by the bat, and the height reached are crucial. The initial speed is 40 meters per second, and the bat adds 70 Joules of kinetic energy to the ball. The mass of the baseball is 0.14 kg, and the total height reached is 26.2 meters. The gravitational acceleration is confirmed to be 9.8 m/s². Using the Conservation of Energy principle, the final speed can be calculated by considering both the initial kinetic energy and the work done by the bat.
Le
Hi,
Would you please help me with this problem.
A batter hits a baseball 1.2 meters above the ground (the point of impact). The ball has mass m = .14g
Speed V(initital) before the ball hit the bat is 40 meters/sec. The bat does 70 Joules of work on the ball in hitting it. Find the speed of the ball after the ball leaves the bat. The ball reaches a height of 25 meters above the point of impact.
Thank you,
Le
 
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Speed V(initital) before the ball hit the bat is 40 meters/sec.

In which direction? Down, I presume? And also, I assume that air resistance is negligible.

From the data given, I just need the mass of the baseball, the gravitational acceleration, and the height it reached. Then, just use the Conservation of Energy. The part before the ball being whacked by the bat is not important at all.
 
[/quote]The part before the ball being whacked by the bat is not important at all.[/quote]

Yes, it is. The problem says " The bat does 70 Joules of work on the ball in hitting it." and so ADDS 70 Joules to the kinetic energy the ball already had. You use the speed of the ball before the hitting the bat to find its kinetic energy then and add 70 Joules.
 
Originally posted by Pandemonium
I just need the mass of the baseball, the gravitational acceleration, and the height it reached.
Isn't acceleration due to gravity 9.8ms^2?

And he says that the mass is .14kg and the height reached is 25m + 1.2m
 
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