Question

Estimate the average force that a baseball pitcher's hand exerts on a 0.145-kg baseball as he throws a 40-m/s pitch. Assume the pitcher accelerates the ball through a distance of 3.0 m, from behind his body to where the ball is released and ignore the air resistance.

Answer 1

Answer:

38.67 N

Explanation:

v = Final velocity = 40 m/s

u = Initial velocity

m = Mass of ball = 0.145kg

s = Displacement of ball = 3 m

Equation of motion

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{40^2-0^2}{2\times 3}=\frac{800}{3}\ m/s^2$

F=ma

$\\\Rightarrow F=0.145\times \frac{800}{3}\\\Rightarrow F=38.67\ N$

∴ Average force that a baseball pitcher's hand exerts on the ball is 38.67 N

Answer 2

Answer:

Force, F = 38.66 N

Explanation:

It is given that,

Mass of the baseball, m = 0.145 kg

Initial speed of the baseball, u = 0

Final speed of the baseball, v = 40 m/s

Distance covered, d = 3 m

Let a is the acceleration of the baseball. It can be calculated using the third equation of motion as :

$a=\dfrac{v^2-u^2}{2d}$

$a=\dfrac{(40)^2}{2\times 3}$

$a=266.66\ m/s^2$

Let F is the average force that a baseball pitcher's hand exerts on a baseball. Using the second equation of motion to find it :

$F=m\times a$

$F=0.145\ kg\times 266.66\ m/s^2$

F = 38.66 N

So, the average force a baseball pitcher's hand exerts on a baseball is 38.66 N. Hence, this is the required solution.