# 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.

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.