Biomechanics Laboratory CProjectile Motion(10 pts)Part 1: Perform calculations to determine angle, speed, and maximum height achieved duringthe 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. Fillin the time in the air for each jump and calculate and average.Measured conditionsJump 1Jump 2Jump 3AverageDistance Sh(m)Takeoffheight (m)ttot (s)ttot (s)ttot (s)ttot (s)0.9500.250.240.240.25Type answers into spaces and table below. Please indicate which equation (1-5) was used foreach calculation below and make sure all units are correct and are indicated: (Note: use only theaverage ttotto perform your calculations).Calculate time up (tup) & time down (tdown):tdown& tup(take-off height and landing height are the same):tdown= tup= ½ (0.25) = 0.125Calculate vertical takeoff velocity (vv):Vfinal= Vinitial+ a t0 = Vinitial + (0.125 sec) (-9.81 m/s^2)0 = Vinitial –(1.22625)Vinitial =1.23 m/sCalculate horizontal takeoff velocity (vh):sh= vhttot(0.95) = Vh (0.25)Vh = 3.8 m/sCalculate peak height of the jumps = vinitialt + ½ a t^2s= 0 + ½ (-9.81 m/s2)(0.125)2s= 0 + ½ (-9.81)(0.015625)s= 0 + (-0.76640625)s= 0.76640625 ms= 0.77 mCalculate angle at take-off & the magnitude of velocity (vmag): (hint: use trigonometry, notequations 1-5)Arctan (Vi/Vh)= xArctan(1.23 m/s/3.8m/s)= xx= 17.93594832 degreesAngle at takeoff= 17.94 degrees**Arctan (opposite/adjacent)Table 2: Part 1 calculation results.Sh(m)Takeoffheight (m)ttot(s)vv(m/s)vh(m/s)Peakheight (m)vmag(m/s)Angle(deg)0.9500.1251.233.80.0773.99417.94Part 2: How does modifying angle, speed (vmag), and take-off height impact horizontal distancetraveled?Case #1: Use calculations from nominal case for take-off height, and angle, while increasingspeed (vmag) by 50%.Vmag = 6 m/s (now)Take-Off height = 0 mAngle = 17.74 degreesVv = sin (17.74 degrees) * 6Vv = 1.83 m/sVh = cos (17.74 degrees) * 6Vh = 5.7 m/sVmag = √(1.832) + (5.72)Vmag = 6 m/sTotal time:Vf = Vi + a * t0 = 1.83 + (-9.81) (t)-1.83 / - 9.81 = tT = 0.19 secondsT total = (0.19) * 2T total = 0.38 secondsHorizontal Distance TraveledSh = Vh * T totalSh = (5.7) * (0.38)Sh = 2.16 metersPeak Heights = vi t + ½ a t^2s = (1.79) (0.182) + (0.5) (-9.81) (0.182^2)s = 0.32578 + (-0.16247)s = 0.163 mCase #2: Use calculations from nominal case for take-off height, and speed (vmag), whileincreasing takeoff angle by 50%.Angle = 26.91 degreesVv = sin (theta) (V take off)Vv = sin (26.91) (3.994 m/s)Vv = 0.45259 * (3.994)Vv = 1.81 m/sVh = cos (theta) (V takeoff)Vh = cos(26.91) * (3.994)Vh = 0.891718 * (3.994)Vh = 3.56 m/sTime to Peak Vertical PositionVfinal = 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 sect(up) = t(down)T (down) = 0.184Total TimeTotal time = 0.184265 + 0.184265t(total) = 0.369 secVertical Distance from Initial to Peak PositionS(up) = Vi + ½ at^2 (up)S(up) = (1.80764 m/s) + ½ (-9.81) ((0.18426^2)S(up) = (1.80764 m/s) + (-0.166533)S(up) = 1.6411 m/sHorizontal DisplacementSh = Vh * t(total)Sh = 3.56152 * 0.36853Sh = 1.31 mCase #3: Use calculations from nominal case for speed (vmag), and takeoff angle, while increasingthe takeoff height to 0.20m.Vmag = 4 m/sTakeoff Angle = 17.74 degreesTake Off Height = 0.20 mVv = sin (17.74 degrees) * 4Vv = 1.22 m/sVh = cos (17.74 degrees) * 4Vh = 3.81 m/sVmag = √(1.222) + (3.812)Vmag = 4 m/sTime to Peak Vertical PositionVf = Vi + a * Tup0 = 1.22 + (-9.81)*Tup-1.22/(-9.81) = TupTup = 0.124 secVertical Position from Initial Position to Peak PositionSup = Vi + (0.5) a * Tup ^2Sup = (1.22) + (0.5)(-9.81)*(0.124^2)Sup = 1.14 mVertical Distance from Peak Position to Landing PositionVf^2 = Vi^2 + 2a * SvVf^2 = 0^2 + 2(-9.81)(1.34)Vf^2 = -26.29Vf = -5.13 m/sTime from Peak to LandingVf = Vi + a* t-5.13 = 0 + (-9.81) (t)-5.13 / (-9.81) = tTdown = 0.523 secTtotalTup + Tdown = Ttotal0.124 + 0.523 = TtotalTtotal = 0.647 secHorizontal DisplacementSh = Vh * TtotalSh = 3.81 * 0.647Sh = 2.47 mCalculate horizontal distance traveled (Sh), vertical velocity (vv), horizontal velocity (vh), andpeak height for each case following instructions for each case, listed below. Fill answers intotable 2. (Note: be consistent with decimal places)Table 3: Part 2 resultsSh(m)Takeoffheight (m)ttot(s)vv(m/s)vh(m/s)Peakheight (m)vmag(m/s)Angle(deg)Case#12.16 m00.38sec1.83m/s5.70m/s0.163 m6 m/s17.74degreesCase#21.31 m00.369sec1.81m/s3.56 m/s1.6411m/s6 m/s17.74degreesCase#32.47 m0.20 m0.647sec1.22m/s3.81 m/s1.34 m/s4 m/s17.74degreesPart 2 Questions to be answered:1. Based on your results, what aspect of the long jump is most important for obtaining themaximum horizontal distance (max sh)?a. The aspect of the long jump that is most important for obtaining the maximumhorizontal distance is speed of release. Long jumpers make the sacrifice of givingup a higher optimal angle in order to get their maximum speed. The speed ofrelease is going to increase, along with the horizontal velocity, and the time in theair. This is because the long jumper is still using some positive number (not zero)angle of release. The two working together will increase the athletesdisplacement, which makes the speed of their release the most important factorwhen achieving their maximum horizontal displacement.2. Based on your previous answer, what instructions might you give a long jumper tomaximize horizontal displacement? Why?a. Instructions that I would give to a long jumper to maximize their horizontaldisplacement is to keep their speed while running on the track high. The fasterthat they are able to have run then the more push off they will receive when theybegin to jump. I would also recommend not slowing down when they get close tothe board. Another aspect that is important for a long jumper is to maintain closeto 20 degrees take off angle release. This will help maximize their jump.Part 3 – Throwing simulationIn throwing, increasing the angle of release will affect the height of release and the speed ofrelease. In Part 3, you will use data collected on an athlete throwing the shot put usingdifferent angles of release. Using an online projectile motion calculator:http://www.walter-fendt.de/html5/phen/projectile_en.htmpredict flight characteristics for the following actual shot put performances that weredetermined by videotaping the athlete putting the shot at five different angles of release (for vvand vhprovide the initial velocities; you