Posted by Sierra on Sunday, November 21, 2010 at 12:53am.
Calculate the loss of total mechanical energy (potential + kinetic) from release until it reaches the basket.
(1/2)M*V1^2 + M*g*H1 = (1/2)MV2^2 + M*g*H2 + (E lost)
Elost = 0.700{[(1/2)(7.7)^2-(3.1)^2] + (9.8)(2.00 - 3.00)} = ____
I consider it unlikely that a basketball would slow down that much due to air resistance, but those are the numbers you were given.
10.5 J = frictional work lost
Initial KE = 20.7 J
Something is phoney here.
Related Questions
AP PHYSICS - A basketball player makes a jump shot. The 0.700 kg ball is ...
AP PHYSICS - A basketball player makes a jump shot. The 0.700 kg ball is ...
AP PHYSICS PLEASE HELP - A basketball player makes a jump shot. The 0.700 kg ...
physics - A basketball player makes a jump shot. The 0.650-kg ball is released ...
physics - A basketball player makes a jump shot. The 0.581 kg ball is released ...
physics - A basketball player makes a jump shot. The 0.585 kg ball is released ...
Physics - A basketball player makes a jump shot. The 0.700 kg ball is released ...
Physics - Trailing by two points, and with only 1.00 s remaining in a basketball...
PHYSICS - Assume: A 78 g basketball is launched at an angle of 46.7 ◦ ...
Physics help!! - Assume: A 78 g basketball is launched at an angle of 56.8&#...
For Further Reading