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A baseball pitcher throws a curveball with an initial velocity of 65 mi/h and a spin of 6500 r/min about a vertical axis. A baseball weighs 0.32 lbf and has a diameter of 2.9 in. Using the data of Fig. P7.108 for turbulent flow, estimate how far such a curveball will have deviated from its straight-line path when it reaches home plate 60.5 ft away. a. approximate solution lb.s/ft2 ax=-22 ft/s2 and x=60.5 ft => x=1/2axt2+v0t => t=0.69 sec => v= axt+v0=79.8 ft/s V2 R ave ay ( 95 79.8 2 ) x 60.5 o 2 898.9 ft; sin 1 sin 1 3.86 8.5 R 898 . 9 y R1 cos 898.91 cos 3.86 2.03 ft b. exact solution Ft mat CD D 2 1 dV (t ) dV ACD 1 ACD AV 2 (t ) m 2 dt t Const 2 dt V 2m V 2m 4 0.00238 slug / ft 3 W 0.32 1 1 m 0.0099 0.00243t 0.0105 or : V g 32.2 0.00243t 0.0105 V CD 0.44 mi ft V (t 0) 65 95 h s 1 V 2 (t ) 2m [1 (dy / dx) 2 ]3 / 2 2m Fn man C L AV 2 (t ) m r 2 2 2 r ACL d y / dx ACL A 0.046 ft 2 dy / dx [1 2 ]3 / 2 2m d ACL ACL dx x Const 2 3/ 2 2 d / dx ACL [1 ] 2m 2m 1 ( x 0) dy / dx( x 0) 0 1 2 ACL 2m dy / dx 2 x 2 1 dy / dx ACL x 2m ACL 2 x 2m ACL 2 ACL 2 2 1 x dy / dx dx x dy 2m 2 2 m 1 ACL x 2 2m 2 2m y ACL 2 ACL 2 2 1 x Const 2m 2m y ( x 0) 0 y AC L at x=60.5: y=1.722 ft ACL 2 2 2m 1 x 2m ACL 2