Question

STANTON WHY NOT Recently, MLBs Statcast also started keeping track of launch angle and projected distance. Verify the projected distance of 434ft corresponding to exit velocity of 109.4mph and 27° launch angle using our projectile motion model. Discuss any discrepancies. tifferences belvieen ar answer and Phis ner Stanton is 66", so you may assume that the ball's initial position is 2'9" (level of Stanton's thighs). EXIT VELOCITY 109 4 LAUNCH ANGLE PROJECTED HR DISTANCE 43 STATCAST aws 27 STANTON EXIT VELOCITY tLIPPA LAUNCH ANGLE 36 109.4 mph = 160, 453 FtIs >> オGeneral accelero-fion funcio a"(t)二<O,-32 ft/ 3242 21 G-142964 C2-72.344 V142.-32++ 72. 니537 niHau Velocity constonts:v sin 27。=045粥 Tw49 "(t) =</42.9044,-kt у~ 72.844 72.844 +16a453 : 27° =0.gn 100-lø0.453 Cos 929 0, 2.9F1) V Co) =<142.964, 72.544> < o, 100, 453> scの=

Please answer the entire question and show all your work (please). Discuss the discrepancies as well. Thank you!

Answer 1

Say at d = 0 x = x_0 & y = y_0 from question x_0 = 0 2y_0 = 2'.9'' = 2.75ft u = 109.4 mph u = 160.453 ft/s say ball hit the ground at to d_f then in y axis y = y_0 + u_y t_f + 1/2 c_y t^2_f = 0 u_y = u sin(27 degree) = 49.67 mph = 72.849 ft/s 2g = -32.2 ft/s^2 therefore 2.75 + 72.849 t_f - 16.1 t^2_f = 0 solving for t_f t_f = 4.56 sec range in t = t_f time x = u_x t_f = u cos 27 degree times 4.56 x = 651.919 ft which is greater than 434ft, The reason for this is that in real time events ball also focus a drag force from environment which is not included in this model hence yielding a higher value than observed one.