Conservation of motion and basketball shot

[xdctassonx]

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.

 

[phyzguy]

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.

 

[xdctassonx]

Got thanks! I actually just calculated the balls maximum height and then treated it as a free fall problem.

 

[phyzguy]

Glad to help. Don't forget the Thanks button!

 

[Nickie]

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!