How much force did the catcher have to use to stop the ball

Question

a bassball pitche throw a 145 gram fastball at 90mph if the catchmet glove moves back 10 cm while catching the ball 

Answer 1

Use the kinematic relation

 

v_f² - v_i² = 2ad

 

v_f = final speed of ball = 0

v_i = initial speed of ball = 90 mph 

d = distance over which it accelerated to a full stop = 10 cm = 0.1 m

a = acceleration = ?

m = mass of ball = 145 g = 0.145 kg

F = force = ma = ?

 

0² - v_i² = 2ad

 

a = -v_i²/(2d)

 

Convert v_i = 90 mph to m/s

 

90 miles/hour = 90(1600 m)/(3600 s) = 40 m/s

 

a = -v_i²/(2d) = -(40 m/s)²/[2(0.1 m)] = -8000 m/s²

 

The minus sign means it is really a deceleration, opposing the direction of motion.

 

The magnitude of the force is

 

F = m|a| = (0.145 kg)(8000 m/s²) = 1160 N