Basketball player jump physics problem

[quicknote]

Would somebody be able to go over my answers to this question. There are a couple of them that I am not sure of. Thank you!
A 100kg baseketball player can leap straight up in the air to a height of 80cm. You can undersdand how by analyzing the situation as follows:
a. The player bends his legs until the upper part of his body has dropped by 60cm, then he begins his jump. Draw separate free body diagrams for the player and for the floor as he is jumping but before his feet leave the ground.
ON PLAYER
There are only two forces when his feet are above to leave the ground, Fn and Fg.
ON FLOOR
The force pointing down is his weight plus ma and using Newton's third law, an equal but opposite reaction is pushing up on him
b. Is there a net force on the player as he jumps (before his feet leave the ground)? How can that be? Explain.
I'm not totally sure on this one:
Before he leaves the ground there isn't a net force because he hasn't actually moved. His feet are still in the same place on the ground. An net force pointing the positive y direction will appear when his feet leave the ground, which pulls him up.
c. WIth what speed must the player leave the ground to reach a height of 80cm?
$$v_{2}^2 = v_{1}^2 + 2ad$$ where d = 0.8 m
Therefore $$v_{2}^2 = 3.96 m/s$$
d. What was his acceleration, assumed to be constant as he jumped?
$$3.96^2 = v_{1}^2 + 2a(0.6)$$
therefore a = 13.068 m/s^2
e. Suppose the player jumps while standing on a bathroom scale that reads in Newtons. What does the scale read before he jumps, as he is jumping and after his feet leave the ground?
Before his jump his weight is shown 100 kg. In Newtons it's 980N
As he is jumping F = ma = 1306N (I'm not sure about this one...Does the acceleration of 13.068 m/s^2 already take into account the affect of g on the player? or do we have to add the affect of gravity as well?)
After his feet leave the ground the scale reads 0N

 

[Doc Al]

A 100kg baseketball player can leap straight up in the air to a height of 80cm. You can undersdand how by analyzing the situation as follows:
a. The player bends his legs until the upper part of his body has dropped by 60cm, then he begins his jump. Draw separate free body diagrams for the player and for the floor as he is jumping but before his feet leave the ground.
ON PLAYER
There are only two forces when his feet are above to leave the ground, Fn and Fg.

Sounds good.

ON FLOOR
The force pointing down is his weight plus ma and using Newton's third law, an equal but opposite reaction is pushing up on him

Careful. Newton's third law says that whatever force the floor exerts on the player (which is Fn), the player must exert the same force (but opposite) on the floor.

b. Is there a net force on the player as he jumps (before his feet leave the ground)? How can that be? Explain.
I'm not totally sure on this one:
Before he leaves the ground there isn't a net force because he hasn't actually moved. His feet are still in the same place on the ground. An net force pointing the positive y direction will appear when his feet leave the ground, which pulls him up.

Just because his feet don't move doesn't mean there is no acceleration of his center of mass. Compare this with your answer to part (d).

c. WIth what speed must the player leave the ground to reach a height of 80cm?
$$v_{2}^2 = v_{1}^2 + 2ad$$ where d = 0.8 m
Therefore $$v_{2}^2 = 3.96 m/s$$

OK.

d. What was his acceleration, assumed to be constant as he jumped?
$$3.96^2 = v_{1}^2 + 2a(0.6)$$
therefore a = 13.068 m/s^2

Assuming (as you are expected to) that the player's center of mass rises 0.6 m, then this is OK. (Reconsider your answer to part b.)

e. Suppose the player jumps while standing on a bathroom scale that reads in Newtons. What does the scale read before he jumps, as he is jumping and after his feet leave the ground?
Before his jump his weight is shown 100 kg. In Newtons it's 980N

OK.

As he is jumping F = ma = 1306N (I'm not sure about this one...Does the acceleration of 13.068 m/s^2 already take into account the affect of g on the player? or do we have to add the affect of gravity as well?)

Yes, you have to take gravity into account. Newton's 2nd law says that the net force equals ma. As you state above, there are two forces acting on the player--the scale will read the normal force.

After his feet leave the ground the scale reads 0N

OK.

 

[quicknote]

.

Your answers are mostly correct, but here are some clarifications:

a. The player's weight (mg) and the normal force (Fn) are both acting on the player when his feet are about to leave the ground. The only difference is that Fn is pointing upwards and mg is pointing downwards. This is because the player is standing on the ground and the ground is pushing up on him with an equal and opposite force (Fn) according to Newton's third law. So, the free body diagram for the player would show both Fn and mg acting on him. Similarly, the free body diagram for the floor would show both Fn and mg acting on it.

b. Before the player's feet leave the ground, there is a net force acting on him. The net force is the vector sum of the forces acting on him, which in this case is Fn - mg. This net force is pointing upwards and is what causes the player to accelerate upwards. Even though the player's feet haven't left the ground yet, he is still accelerating upwards due to this net force.

c. Your calculation for the speed is correct, but you should also include the acceleration due to gravity (g) in your calculation. The equation should be v2^2 = v1^2 + 2ad + g, where d = 0.8m and g = 9.8m/s^2. This will give you a slightly different answer for the speed.

d. Your calculation for the acceleration is correct. Just remember to include the acceleration due to gravity (g) in your calculation as well.

e. The scale will read 980N before the player jumps, because that is the player's weight. As he is jumping, the scale will read a higher value, because the player is accelerating upwards due to the net force. This value should be higher than 1306N, because it should include the weight (mg) and the normal force (Fn). After his feet leave the ground, the scale will read 0N, because there are no forces acting on the player anymore.

Overall, your understanding of the physics involved in this situation is good. Just remember to consider all the forces acting on the player and to include the acceleration due to gravity in your calculations. Keep up the good work!