Professional baseball pitchers deliver pitches that can reach the blazing speed of 100 mph (miles per hour). A local team has drafted an up-and-coming, left-handed pitcher who can consistently pitch at 42.24 m/s (94.50 mph).
A. Assuming a pitched ball has a mass of 0.1420 kg and has this speed just before a batter makes contact with it, how much kinetic energy does the ball have?
B. How high would the ball need to be dropped from to attain the same energy (neglect air resistance)?

Assuming no air resistance, once released by the pitcher, the speed must keep constant through all the trajectory, so the kinetic energy of the ball can be expressed as follows:

[tex]K = \frac{1}{2}*m*v^{2} = \frac{1}{2}*0.142 kg*(42.24m/s)^{2} = 126.7 J (1)[/tex]

B)

Neglecting air resistance, total mechanical energy must be the same at any point, so, if we choose the ground level as the zero reference level for the gravitational potential energy, and assuming that the ball attains this kinetic energy just before striking ground, this value must be equal to the gravitational potential energy just before be dropped, so we can write the following equality:

[tex]U_{o} = K_{f} = 126. 7 J (2)[/tex]

⇒ m*g*h = 126. 7 J

[tex]h = \frac{K_{f}}{m*g} = \frac{126.7J}{0.1420kg*9.8m/s2} = 91.1 m (3)[/tex]