Question

A baseball outfielder catches a fly ball traveling 25.0 m/s with his gloved hand. When he catches the 0.140 kg baseball, the outfielder’s hand recoils by 10.0 cm. How much force did the outfielder’s hand exert in catching the ball?

Answer 1

The force exerted by the outfielder's hand in catching the baseball can be calculated using the work-energy principle, equating the work done by the force to the change in kinetic energy of the ball brought to rest.

Step-by-step explanation:

To determine the force exerted by the outfielder's hand in catching the baseball, we can use the work-energy principle. The work done by the force is equal to the change in kinetic energy of the ball.

The initial kinetic energy (KEi) of the ball is given by ½mv², where m is the mass of the ball and v is its velocity just before being caught. Since the ball comes to a stop in the hand, its final kinetic energy (KEf) is 0. The work W done by the hand is the force exerted multiplied by the distance over which this force acts (displacement).

Therefore, we have:

½mv² = Fd

Substituting the given values (½ × 0.140 kg × (25.0 m/s)² = F × 0.10 m), we can solve for the force F.

Answer 2

10cm=0.1m

s=0.1
u=25
a=?
v=0

v ²=u ²+2as
0=25 ²+2a(0.1)
0=625+ 0.2 a
-625=0.2a
a=-3125 m/s²

F=ma
F=(0.14)(-3125)
F=-437.5N

F=437.5