Question

A 110-kg basketball player can jump a vertical distance of 1.3 meters (approx. 4 feet). For this entire problem, assume that the athlete has NO horizontal motion. (Ignore air resistance.) 6. (2 pts.) What must be the player's initial vertical velocity as he leaves the ground? A. 3.4m/s B. 5.0 m/s D. 9.2 m/s E II m/s C. 7.6 m/s 7. (I pt.) At the moment the athlete reaches his maximum height, which of the following is/are equal to zere? A. his velocity B. his speed C. his acceleration D. A & B only E all of the above s (2 pts.) How mach total time elapses during his entire jump, from take-off to landing? A. 0.78s B. 10 s C. 13s D. 1.6s E. 1.9s smale, 734 athete mps with the senitial velocity as the 10-lg ahicte does. Which ose of the following is TRUE? A-The 75-kg-lete will reach a preater maximum height. B. The 75-kg athlete will reach a lower maximum beight. C. The 75-kg athlete will be in the air for a longer period of time before landing. D. The 73-kg athlete will be in the air for a shorter period of time before landing. E None of the above. + Six different cars move along one dimesional tacks. The following sis graphs represest their VELOCITIES as functions of time: 13) 10. (l pt) Which of the above carto) ha ant speed A. (1) only B. (2) only C. (6) only D. (2) & (6) only 11. (I pt) Which of the above car(s) has a constant negative acceleration? A. (4) only B. (5) only C. (6) only D. (3) & (6) only 12.(l pt.) Which of the above carfs) is slow ing down? D. (5) & (6) only A. (4) only B. (5) only C. (6) oaly E (4), (5) & (6) Ar 13. (2 pts.) Which of the above carts) has a displacement graph roughly like this one: D. (5) & (6) only A. (4) only B. (5) only C. (6) only

Answer 1

n = 1.3 m, m = 110kg At highest point v = 0 for upward motion a = -g = -9.8 m/s^2 Then from v^2 = u^2 + 2ah 0 = u^2 - 2 times 9.8 times 1.3 or u = squareroot 2 times 9.8 times 1.3 u TildeEqual 5.0 m/s (B) At maximum height velocity & speed both are zero = > Option (D) for upward motion v = u + at = > 0 = 5 - 9.8t or t = 5/9.8 = > t = 0.5 sec Therefore total time T = 2t = 2. Times 0.5 T = 1sec (B) Now m_2 = 75Kg & equation of motion are v = u + at h = ut + 1/2 at^2 & v^2 = u^2 + 2ah Equations do not depend on the mass whatever be the mass, h & T will depend only on initial velocity & acceleration h & T will be same in both of the cases Option (E)