Class Management Help HW 6 Begin Date: 6/17/2018 12:00:00 AM -- Due Date: 7/31/2
ID: 1731781 • Letter: C
Question
Class Management Help HW 6 Begin Date: 6/17/2018 12:00:00 AM -- Due Date: 7/31/2018 11:59:00 PM End Date: 8/16/2018 11:59:00 PM (20%) Problem 5: A basketball player jumps straight up for a ball. To do this, he lowers his body 0.325 m and then accelerates through this distance by forcefully straightening his legs. This player leaves the floor with a vertical velocity sufficient to carry him 0.895 m above the floor. For this problem, use a coordinate system where up is positive 33% Part (a) Calculate his velocity when he leaves the floor in m/s Grade Summary 0% 100% Potential tanO) T coS asin) atan acotan)sinhO coshOtanh0 cotanhO Degrees O Radians sin Submissions Attempts remaining: 10 (5% per attempt) detailed view cotan 0 Submit Hint I give up! Hints: 3% dcduction per hint. Hints remaining: 4 Feedback: 3% deduction per fccdback. 33% Part (b) Calculate his acceleration in m/s2 while he is straightening his legs. He goes from zero to the velocity found in part (a) in a distance of 0325 m 33% Part (c) Calculate the magnitude of the force he exerts on the floor to do this, in Newtons, given that his mass is 116 kgExplanation / Answer
Here we have to find the velocity when he leaves i.e the initial velocity (V1)
As we can see that the basketball player is like a projectile
Since he reaches a maximum height of 0.895 m (h)
As we know that
V2^2 = V1^2 - 2*g* (Height)
V2 = 0 (Final Velocity)
0 = V1^2 - 2*g*h
V1 = sqrt(2*9.8*0.895)
V1 = 4.18 m/s ( the velocity when he leaves floor)
2 ( ), 0 2 0 2 v = v ? g y ? y with 0.900m, and 0 m/s. y ? y0 = v = Solving for the initial velocity gives: [2 ( )] [2(9.80 m/s )(0.900 m)] 4.20 m/s