Chapman holds the record for fastest pitch at 47 m/s. He threw a pitch at an elevation angle of 15 degrees, and the maximum altitude reached by the baseball was calculated. If the same 47 m/s pitch was thrown at a 45 degree angle, the range was calculated. The maximum altitude was found to be 112.7 m, and the range at 45 degrees was calculated to be 225.41 m.

1. Aroldis Chapman, relief pitcher for the New York Yankees baseball team, holds the record for
the fastest baseball ever thrown. Chapman's record pitch velocity is v 0?= 47.0 m/s. Chapman
threw a baseball with a velocity of v 0? into the outfield at an elevation angle ? =15°. Calculate
the h , the maximum altitude of the baseball after Chapman threw it.
Aroldis Chapman, relief pitcher for the New York Yankees baseball team, holds the record for
the fastest baseball ever thrown. Chapman's record pitch velocity is v 0?= 47.0 m/s. If Chapman
had thrown the baseball with a velocity of v 0? into the outfield at an elevation angle ? =45°,
then calculate its range R .
Solution
Given
the velocity (initial) is v0 = 47 m/s
the angle of projection is theta = 15 degrees
the range of the baseball is
R = v0^2 sin2theta/g
substituting the values
R = 47^2 sin(2*15)/9.8 m
R = 112.70 m
and the time of flight is
T = 2*v0 sin theta/g
substituting the values
T = 2*47 sin15/9.8 s

2. T = 2.48255 s
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the angle of projection is 45 degrees the range will be maximum
R = v0^2 sin 2 theta/g
R =47^2 sin(2*45)/9.8 m
R = 225.41 m