A baseball pitcher throws a fastball that leaves his hand at a speed of 36.0m/s. The mass of the baseball is 0.145kg. Ignore air resistance. How much work has the pitcher done on the ball in throwing it?
I dunno what to use
$\displaystyle W = \frac{1}{2}m{v_{2}}^2 - \frac{1}{2}m{v_{1}}^2 $
help please
the answer is 94.0 Joules
The work done is equal to the change in KE of the ball. In this case the initialOriginally Posted by ^_^Engineer_Adam^_^
A baseball pitcher throws a fastball that leaves his hand at a speed of 36.0m/s. The mass of the baseball is 0.145kg. Ignore air resistance. How much work has the pitcher done on the ball in throwing it?
I dunno what to use
$\displaystyle W = \frac{1}{2}m{v_{2}}^2 - \frac{1}{2}m{v_{1}}^2 $
help please
the answer is 94.0 Joules
KE is zero (zero speed), and the final KE is $\displaystyle \frac{0.145\times 36^2}{2}$
RonL
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