A basketball player jumps straight up for a ball. To do this, he lowers his body 0.320 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 0.880 m above the floor. (a) Calculate his velocity (in m/s) when he leaves the floor. (Enter a number.) m/s (b) Calculate his acceleration (in m/s) while he is straightening his legs. He goes from zero to the velocity found in part (a) in a distance of 0.320 m. (Enter a number.) m/ s 2 (c) Calculate the force (in N) he exerts on the floor to do this, given that his mass is 102 kg. (Enter a number.) NPage No. Date : Clarification Paint Ca ) Now , At the top of V = Q player's movement player will come at gest . T R= O.prom Hence , velocity, u = 0 Gravitational acceleration, 9 will act against the direction of motion. Hence it will decelerate the player. Distance travelled from bottom to top and using mation equation , Uz = 42 - 29 6 we have found the result . Paint ( b ) So, the acceleration achieved from snest Due to to final velocity Lowenmy 4.153 m/s (rise . velocity Body At After at which he leaves the bottom ) Rest acceleration will be in the distance uso velocity = 4:153 m /s of 0 320 m