CP BIO? A Standing Vertical Jump. Basketball player Darrell Griffith is on record as attaining a standing vertical jump of 1.2 m (4 ft). (This means that he moved upward by 1.2 m after his feet left the floor.) Griffith weighed 890 N (200 lb). (a) What was his speed as he left the floor? (b) If the time of the part of the jump before his feet left the floor was 0.300 s, what was his aver-age acceleration (magnitude and direction) while he pushed against the floor? (c) Draw his free-body diagram. In terms of the forces on the diagram, what was the net force on him? Use Newton’s laws and the results of part (b) to calculate the average force he applied to the ground.

Solution 39P Step 1 : In this question, we need to find Speed with which Darrell griffith left the floir v Average acceleration ( magnitude and direction ) while he pushed against the floor Draw his body diagram and find net force on him Average force applied by player in ground Step 2 : Let us consider the data given Maximum height reached h = 1.2 m Mass of player m = 890 N or 200 lb Time taken for part of jump t = 0.300 s Step 3 : We shall find the initial velocity of the vertical jump using 1mv = mgh 2 Solving this for velocity we get v = 2gh Gravitational force g = 9.8 ms2 Substituting the values we get v = 2 × 9.8 m/s × 1.2 m v = 23.52 m /s 2 v = 4.849 m/s Hence we have obtained velocity as 4.849 m/s Step 4 : Let us find the acceleration of the player We have initial velocity as v = 0 i/s And final velocity as v = 4f489 m/s Time is given as t = 0.300s Thus we have use relation v a = t Substituting the values we get a = vfvi t 4.489 m/s 0 m/s a = 0.300 s a = 16.16 m/s 2 Hence we have acceleration as a = 16.16 m/s 2