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Conservation of motion and basketball shot

  • Thread starter xdctassonx
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  • #1
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.
 

Answers and Replies

  • #2
phyzguy
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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
phyzguy
Science Advisor
4,506
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Glad to help. Don't forget the Thanks button!
 

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