Biomechanics Laboratory C Projectile Motion 10 pts Part 1 Perform calculations to determine angle speed and maximum height achieved during the jump based on information obtained from data collection i e time and distance Table 1 Measured conditions We have measured and set the distance of the jump for you Fill in the time in the air for each jump and calculate and average Measured conditions Jump 1 Jump 2 Jump 3 Average Distance Sh m Takeoff height m ttot s ttot s ttot s ttot s 0 95 0 0 25 0 24 0 24 0 25 Type answers into spaces and table below Please indicate which equation 1 5 was used for each calculation below and make sure all units are correct and are indicated Note use only the average ttot to perform your calculations Calculate time up tup time down tdown tdown tup take off height and landing height are the same tdown tup 0 25 0 125 Calculate vertical takeoff velocity vv Vfinal Vinitial a t 0 Vinitial 0 125 sec 9 81 m s 2 0 Vinitial 1 22625 Vinitial 1 23 m s Calculate horizontal takeoff velocity vh sh vh ttot 0 95 Vh 0 25 Vh 3 8 m s Calculate angle at take off the magnitude of velocity vmag hint use trigonometry not equations 1 5 Calculate peak height of the jump s vinitial t a t 2 s 0 9 81 m s2 0 125 2 s 0 9 81 0 015625 s 0 0 76640625 s 0 76640625 m s 0 77 m Arctan Vi Vh x Arctan 1 23 m s 3 8m s x x 17 93594832 degrees Angle at takeoff 17 94 degrees Arctan opposite adjacent Table 2 Part 1 calculation results Sh m Takeoff height m ttot s vv m s vh m s Peak height m vmag m s Angle deg 0 95 0 0 125 1 23 3 8 0 077 3 994 17 94 Part 2 How does modifying angle speed vmag and take off height impact horizontal distance traveled Case 1 Use calculations from nominal case for take off height and angle while increasing speed vmag by 50 Vmag 6 m s now Take Off height 0 m Angle 17 74 degrees Vv sin 17 74 degrees 6 Vv 1 83 m s Vh cos 17 74 degrees 6 Vh 5 7 m s Vmag 1 832 5 72 Vmag 6 m s Total time Vf Vi a t 0 1 83 9 81 t 1 83 9 81 t T 0 19 seconds T total 0 19 2 T total 0 38 seconds Horizontal Distance Traveled Sh Vh T total Sh 5 7 0 38 Sh 2 16 meters Peak Height s vi t a t 2 s 1 79 0 182 0 5 9 81 0 182 2 s 0 32578 0 16247 s 0 163 m Angle 26 91 degrees Vv sin theta V take off Vv sin 26 91 3 994 m s Vv 0 45259 3 994 Vv 1 81 m s Case 2 Use calculations from nominal case for take off height and speed vmag while increasing takeoff angle by 50 Vh cos theta V takeoff Vh cos 26 91 3 994 Vh 0 891718 3 994 Vh 3 56 m s Time to Peak Vertical Position Vfinal Vinitial a t up 0 1 807645 m s 9 81m s 2 t up 1 807645 9 81 t up t up 0 184 sec t up t down T down 0 184 Total Time Total time 0 184265 0 184265 t total 0 369 sec Vertical Distance from Initial to Peak Position S up Vi at 2 up S up 1 80764 m s 9 81 0 18426 2 S up 1 80764 m s 0 166533 S up 1 6411 m s Horizontal Displacement Sh Vh t total Sh 3 56152 0 36853 Sh 1 31 m Vmag 4 m s Takeoff Angle 17 74 degrees Take Off Height 0 20 m Case 3 Use calculations from nominal case for speed vmag and takeoff angle while increasing the takeoff height to 0 20m Vv sin 17 74 degrees 4 Vv 1 22 m s Vh cos 17 74 degrees 4 Vh 3 81 m s Vmag 1 222 3 812 Vmag 4 m s Time to Peak Vertical Position Vf Vi a Tup 0 1 22 9 81 Tup 1 22 9 81 Tup Tup 0 124 sec Vertical Position from Initial Position to Peak Position Sup Vi 0 5 a Tup 2 Sup 1 22 0 5 9 81 0 124 2 Sup 1 14 m Vertical Distance from Peak Position to Landing Position Vf 2 Vi 2 2a Sv Vf 2 0 2 2 9 81 1 34 Vf 2 26 29 Vf 5 13 m s Time from Peak to Landing Vf Vi a t 5 13 0 9 81 t 5 13 9 81 t Tdown 0 523 sec Ttotal Tup Tdown Ttotal 0 124 0 523 Ttotal Ttotal 0 647 sec Horizontal Displacement Sh Vh Ttotal Sh 3 81 0 647 Sh 2 47 m Calculate horizontal distance traveled Sh vertical velocity vv horizontal velocity vh and peak height for each case following instructions for each case listed below Fill answers into table 2 Note be consistent with decimal places Table 3 Part 2 results Sh m Takeoff height m ttot s vv m s vh m s Peak height m vmag m s Angle deg 5 70 m s 0 163 m 6 m s 17 74 degrees 3 56 m s 1 6411 m s 6 m s 17 74 degrees 2 16 m 1 31 m 0 0 Case 1 Case 2 Case 3 0 38 sec 0 369 sec 0 647 sec 1 83 m s 1 81 m s 1 22 m s 2 47 m 0 20 m 3 81 m s 1 34 m s 4 m s 17 74 degrees Part 2 Questions to be answered 1 Based on your results what aspect of the long jump is most important for obtaining the maximum horizontal distance max sh a The aspect of the long jump that is most important for obtaining the maximum horizontal distance is speed of release Long jumpers make the sacrifice of giving up a higher optimal angle in order to get their maximum speed The speed of release is going to increase along with the horizontal velocity and the time in the air This is because the long jumper is still using some positive number not zero angle of release The two working together will increase the athletes displacement which makes the speed of their release the most important factor when achieving their maximum horizontal displacement 2 Based on your previous answer what instructions might you give a long jumper to maximize horizontal displacement Why a Instructions that I would give to a long jumper to maximize their horizontal displacement is to keep their speed while running on the track high The faster that they are able to have run then the more push off they will receive …