Question

The fastest pitched baseball was clocked at 46 m/s. If thepitcher exerted his force (assumed to be horizontal and constant)over a distance of 1.0 m and abaseball has a mass of 145 g.

What force did the pitcher exert on the ball during thisrecord-setting pitch?

Express your answer using two significant figures.

Estimate the force in Part B as a fraction of thepitcher’s weight. A typical pitcher weighs 170 lbs

Answer 1

We can use one of newton's equations of motions to determinethe acceleration of the ball:V_f²=V_i²+2axwhereV_f is the final velocity, V_i isthe initialvelocity, a is the acceleration and x is the distanceAssuming the initial velocity to be zero:a = V_f² / (2x) = (46m/s)² /(2*1m) = 1058 m/s²Therefore, using newton's second law, F=ma, we can determinethe force of the ball:F = ma = .145kg*1058m/s² = 153.41 NThe force of a typical pitchers mass is:F = m*g = 77.11kg*9.81m/s² = 756.45NTherefore, the fraction of the force of the pitchers weightis:[(153.41N) / 756.45N] *100 = 20.28%

Answer 2

We know from the relationv² - u² = 2aSHere,v = 46 m/s,u = 0 m/s and S = 1.0 m.or a = (v² - u²/2S)or a = [(46)² - 0²/(2 * 1.0)]or a = (2116/2) = 1058 m/s²The force exerted by the pitcher on the ball during thisrecord-setting pitch isF = m * aHere,m = 145g = 145 * 10^-3kgor F = 145 * 10^-3 * 1058 = 153.41 NThe weight of the pitcher is W = 170 lbsWe know that1 lbs = 0.4535 kgor W = 170 * 0.4535 = 77 kgor W = 77 * 9.8 = 754.6 NTherefore,the force in part B as a fraction of the pitcher'sweight is(W/F) = (754.6/153.41) = 4.9or W = 4.9 * FTherefore,the pitcher's weight is 4.9 times the force exertedby the pitcher on the ball during his record-setting pitch.