How Is Average Force Calculated When Catching a Baseball?

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To calculate the average force when catching a baseball, the initial speed of the ball (44.7 m/sec) and its mass (0.50 kg) are used to determine kinetic energy, resulting in 4895.32 J. The acceleration is calculated using the kinematic equation, yielding a value of -73.5 m/sec². Applying Newton's second law, the force is calculated as 37.25 N. The negative value indicates the direction of the force is opposite to the ball's motion, which is acceptable in physics. Thus, the average force on the person's hand is 37.25 N in the direction opposing the baseball's initial velocity.
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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.
 
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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.
 
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.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...

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