How Is Average Force Calculated When Catching a Baseball?

[lochs]

I need help with this question as well: If the baseball in example 4.5 is caught by a person whose hand recoils 0.30 m, calculate the average force on the person's hand. (The initial speed of the baseball is 44.7 m/sec and its mass is 0.50 kg.)

This is what i have, but I'm not sure if it's right...

w=mg I'm not even sure if i have to convert that or not
w=(0.50 kg)(9.8 m/sec2)
w=4.9Nthen I'm not sure how to represent kenitic energy on here but here it goes,
KE=(1/2mv)2 ..(squared)
KE= 1/2(4.9 N)(44.7 m/sec)2
KE= 4895.32 J

thats what i have, but I am not sure if i should convert mass to weight if not then go i just put the 0.50 in the KE equation? Any help would be appreciated.

 

[Jameson]

Use Newton's 2nd Law: $$a = \frac{F_{net}}{m}$$ or $$F_{net}=ma$$

To find "a" is the tricky part. You have to use the information they gave you. I would use the kinematic equation:

$$v_{f}^2 = v_{i}^2 + 2ad$$

Vf = 0
Vi = 44.7
d = .3
a = ?

You should be able to solve it from here.

 

[lochs]

Thanks =)
ok for acceleration i got -73.5 m/sec

then i did
F=ma
F=(0.50 kg)(-74.5 m/sec)
F=37.25 N

now my question is can you have negative force, cause i got a negative number ..but i didnt think you could have negative force.