Question

Chapter D3, Problem D3/167 The baseball is traveling with a horizontal velocity of 81 ml/hr just before impact with the bat. Just after the impact, the velocity of the 5.125-oz ball is 127 ml/hr directed at 27° to the horizontal as shown. Determine the x- and y-components of the average force R exerted by the bat on the baseball during the 0.0061-sec impact. Comment on the treatment of the weight of the baseball (a) during the impact and (b) over the first few seconds after impact. 127 mi/hr 27 81 milhr Answer: R=( 1) i

Answer 1

draue the Let us first baseball during the considering all given conditions impulse - momentum voliagsam ļ the impact. (mg) at the = 127 mithy → V2 Rx.at 1.27 Vio 81 milho Ry.at Vector be written as Velocities in the form of component can - 81 2 (127. Cos 27 )i +(127. sin 27) j ALSO Reaction R R (Rx i + Ry ; ) cb (5.125) ub 5.125 02 given 0.0061 sec t F.olt maV using equation we In direction 2 - get m ( Vx2 - Vai) 0 mz W w Rrit [: - (-BL. 5 280 cos27 (57625) 52.80 3600 (127 Rx ) ] 3600 32 0.0061 Here we have done unit conv esion in velocity

467.284 In 17 direction Ryit - (mg.): t n (Vyan sa vy, ) Ryot W (Vy2 - Vyi) + g Ry S280 sim 27 o y got 15:12). (2) 127. 3600 32 S.125 + 16 0.0061 Ry 139.11 ub Hence, R = ( 467.284 + 139.11 3 ) lb Answer a on change mo Comments about weight of baseball If "we neglect weight of baseball in it's weight gives a significant change in Reaction in y direction but in this change can observe 0 from equation the baseball weight negleact If never fall ground which possible Rx be ball (b we On wiu not Hence, кое can not neglect base ball's weight.
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