# Newtons law question (Baseketball player)

• lovemake1
In summary, the conversation discusses calculating the constant force required for a 70.0 kg person to achieve a vertical jump of 0.80m. The first step is to find the required velocity by using the equation vf^2 = vi^2 + 2ad. Then, the acceleration can be found by using the equation a = vf^2 / 2d. However, there is a mistake in using the distance of 0.8m instead of the given 0.2m, resulting in an incorrect answer. The concept of gravity taking over once the feet leave the floor is also clarified.
lovemake1
Question reads: An exceptional vertical jump from rest would raise a person 0.80m off the ground. to do this what constant force would a 70.0 kg person have to exert againts the ground. Assume the person lowers himself by 0.20m prior to jumping and remains in a standing position while in the air.

My progress: well firstly i drew a free body diagram of him on the floor. (Fn[up] and Fg[Down]) and when he is in the air (Fg[Down])
then i tried finding out the velocity and hopefully i can find acceleration afterwards.

i used vf^2 = vi^2 + 2ad
i set accelration as (-9.8m/s^2) because once he is in the air gravitational force pulls him down.

am i going in the wrong direction ? where should i begin this problem. I am also confused why crouching 0.20m prior to jumping would change anything.

please further guide me with this problem. thanks.

You have 2 problems to deal with, first is a freefall problem to find what velocity you need to give the required height. Then you need to find the acceleration required to give you that velocity.

ok, now i understand that you can use the same equation to find the velocity and acceleration by plotting the numbers correctly.

so i found the velocity by using equation vf^2 = vi^2 + 2ad
vf^2 = 0 +2(-9.8)0.8
vf = 3.95m/s
(but question said he crouched 0.2m where does this go ??)using this velocity i sub it back in 3.95 = 0 + 1.6a
a = 2.46m/s^2

with this acceleration i use Newton's 2nd law

Fnet = ma
Fapp - Fg = ma
Fapp = ma +Fg
Fapp = 70(2.46) + 70(9.8)
fapp = 858.2N

But wait,The answer sheet says it should be 3400N or 340N (my teacher writing is messy)
what did i possibly do wrong here?

lovemake1 said:
ok, now i understand that you can use the same equation to find the velocity and acceleration by plotting the numbers correctly.

so i found the velocity by using equation vf^2 = vi^2 + 2ad
vf^2 = 0 +2(-9.8)0.8
vf = 3.95m/s

This looks good to here

(but question said he crouched 0.2m where does this go ??)

using this velocity i sub it back in 3.95 = 0 + 1.6a
a = 2.46m/s^2
Where did this come from, please show us.

Here is where you need to use the .2m crouch. You have a zero inital velocity and the final velocity you have found above. You need to find the acceleration which works with these values.
with this acceleration i use Newton's 2nd law

Fnet = ma
Fapp - Fg = ma
Fapp = ma +Fg
Fapp = 70(2.46) + 70(9.8)
fapp = 858.2N

But wait,The answer sheet says it should be 3400N or 340N (my teacher writing is messy)
what did i possibly do wrong here?[/QUOTE]

ah ha... there was my problem.

corrected answer: Vf^2 = vi^2 + 2ad
3.95^2 = 0 + 2a(0.2)
15.6025 = 0.4a
a= 39m/s^2Fnet = ma
fapp - mg = ma
fapp = ma + mg
fapp = 2730 + 686
fapp = 3400N

i made a mistake using 0.8m for the distance rather than 0.2m
but I am not sure why this is the case. he achieved the velocity of 3.95m/s during his flight.
not just in a mere distance of 0.2m ? can you clarify this concept more clear?

When jumping you can only accelerate upwards as long as your feet are on the floor. As soon as your feet leave the floor, gravity takes over and you now accelerate at g.

As soon as your legs are straight your feet leave the floor. That occurs at .2m.

great, thanks for your help :D

## 1. What is Newton's First Law and how does it apply to a basketball player?

Newton's First Law states that an object at rest will remain at rest and an object in motion will remain in motion at a constant velocity unless acted upon by an unbalanced force. In basketball, this means that a player will continue to move in a straight line at a constant speed unless a force, such as a defender, changes their motion.

## 2. How does Newton's Second Law relate to a basketball player's performance?

Newton's Second Law states that the acceleration of an object is directly proportional to the force applied to it and inversely proportional to its mass. In basketball, this means that a player's acceleration, or how quickly they can change their speed or direction, is affected by the force they apply and their mass. A heavier player may have a harder time accelerating quickly compared to a lighter player.

## 3. How does Newton's Third Law explain the physical interactions between basketball players?

Newton's Third Law states that for every action, there is an equal and opposite reaction. In basketball, this means that when two players come into contact, the force they exert on each other is equal and opposite. For example, when a player pushes off of the ground to jump, the ground pushes back with an equal force, allowing the player to gain height.

## 4. How do Newton's laws apply to shooting a basketball?

When shooting a basketball, Newton's First Law applies as the ball will continue to move in a straight line at a constant speed unless acted upon by a force, such as air resistance. Newton's Second Law applies as the force applied to the ball determines its acceleration and trajectory. Newton's Third Law applies as the force of the player's hand pushing the ball is equal and opposite to the force of the ball pushing back on their hand.

## 5. Can Newton's laws be used to predict the outcome of a basketball game?

While Newton's laws can help explain the physical aspects of basketball, they cannot be used to predict the outcome of a game. Other factors, such as strategy, skill, and teamwork, also play a significant role in the outcome of a basketball game.

Replies
8
Views
1K
Replies
38
Views
3K
Replies
4
Views
3K
Replies
28
Views
3K
Replies
1
Views
1K
Replies
15
Views
3K
Replies
1
Views
7K
Replies
3
Views
2K
Replies
11
Views
2K
Replies
3
Views
2K