Conservation of motion and basketball shot

In summary, a basketball with a mass of 0.624 kg was shot from a height of 1.55 m at a speed of 12.1 m/s. Using the principle of energy conservation, the ball was found to be moving at a speed of 13.26 m/s just before entering the hoop. However, after some calculation errors were corrected, the correct answer was determined to be less than 12.1 m/s.
  • #1
xdctassonx
6
0
A basketball of mass 0.624 kg is shot from a vertical height of 1.55 m and at a speed of 12.1 m/s. After reaching its maximum height, the ball moves into the hoop on its downward path, at 3.05 m above the ground. Using the principle of energy conservation, determine how fast the ball is moving just before it enters the hoop.

v=√(v0^2+2g(y0-y))

I've tried plugging the numbers in an I come up with an answer of 13.26 m/s. However this is incorrect. I believe I might be thinking of where the ball is being shot from differently.

v=√((12.1^2)+2(9.81)(1.55-3.05)

Any help would be appreciated. Thanks, Dominic.
 
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  • #2
xdctassonx said:
A basketball of mass 0.624 kg is shot from a vertical height of 1.55 m and at a speed of 12.1 m/s. After reaching its maximum height, the ball moves into the hoop on its downward path, at 3.05 m above the ground. Using the principle of energy conservation, determine how fast the ball is moving just before it enters the hoop.

v=√(v0^2+2g(y0-y))

I've tried plugging the numbers in an I come up with an answer of 13.26 m/s. However this is incorrect. I believe I might be thinking of where the ball is being shot from differently.

v=√((12.1^2)+2(9.81)(1.55-3.05)

Any help would be appreciated. Thanks, Dominic.

It looks like you have the formula correct, and you have put the numbers in correctly, but you must have made a mistake doing the calculation. You are taking the 12.1^2, adding a negative number, then taking the square root, so the answer needs to come out less than 12.1 m/s. Check the calculation again.
 
  • #3
Got thanks! I actually just calculated the balls maximum height and then treated it as a free fall problem.
 
  • #4
Glad to help. Don't forget the Thanks button!
 
  • #5


Hello Dominic,

I can confirm that your equation is correct and your approach is on the right track. However, there are a few factors that could be affecting your answer.

First, make sure that you are using the correct units for your calculations. In this case, the speed should be in meters per second (m/s) and the height should be in meters (m). This will ensure that your final answer is also in m/s.

Secondly, check your calculations to make sure you are using the correct values for the initial and final heights. The initial height should be 1.55 m, but the final height should be 0.3 m (3.05 m above the ground). This is because the ball is entering the hoop, which is 3.05 m above the ground, and not at ground level.

Finally, double check your calculation for the acceleration due to gravity (g). The standard value for this is 9.81 m/s^2, but if you are using a different value, it could affect your final answer.

Overall, using the principle of energy conservation and the correct values, the final speed of the ball just before it enters the hoop should be approximately 12.54 m/s. I hope this helps clarify any confusion and good luck with your calculations!
 

1. What is the conservation of motion?

The conservation of motion is a fundamental principle in physics which states that the total momentum of a closed system remains constant. This means that in the absence of external forces, the total momentum of the system before and after a collision will be the same.

2. How does the conservation of motion apply to a basketball shot?

When a basketball is shot, the player applies a force to the ball which causes it to accelerate and gain momentum. As the ball travels through the air, it experiences air resistance and other external forces, but the total momentum of the ball and the player remains constant. This means that the ball will continue to move in a straight line at a constant speed until it is either stopped by a force or reaches the basket.

3. Why is the conservation of motion important in basketball?

The conservation of motion is important in basketball because it explains the trajectory and movement of the ball during a shot. It allows players to predict the path of the ball and make accurate passes and shots. It also helps coaches and players analyze and adjust technique in order to improve shot accuracy and consistency.

4. Are there any exceptions to the conservation of motion in basketball?

In basketball, the conservation of motion applies to the ball and the player, but there can be exceptions when external forces are present. For example, if a player is fouled while shooting, the external force of the defender may alter the trajectory and momentum of the ball, causing it to deviate from its original path.

5. How can understanding the conservation of motion improve basketball performance?

By understanding the conservation of motion, players and coaches can make adjustments to their technique and positioning to optimize the momentum and trajectory of the ball. This can lead to improved accuracy and consistency in shooting, passing, and rebounding. It can also help players anticipate and react to the movement of the ball and other players on the court.

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