Question

A baseball is hit at an angle of 40° above the horizontal with an initial speed of Vo=26.4 m/s. At what time does it have a y component of velocity equal to vy=-8.3 m/s? Select one: 0.71 s 1.29 s 1.06 S 2.58 s 3.875

Answer 1

Initial speed of the ball Vo= 26.4m/s Angle O= 40 Final y.com ponent of velocity uy--8-3-0)s uileal y-component of dy ve velocity go = vo sino Ko va ay=-g=-9.8m/5 Uy = gotat -8-3 = 26.4 x sinuó) - (98xt) 9.8x6= 16.96578.3 9.8% = 25.27 t - 25.27 lat 1 Mass IR time t=2.585 y-component of velocity j --83mba of the ball M=2.2 kg Radings of the ball's' Inilal height of the ball h=713m Rading of the loop. R=2.1m Initial tolal l energy of the ball.TE: = PE: +kE:+RKE; Ay 14 at rest Wouls, ke;= RkE:=oj TE: = mgh:= 2+2x9.8* 7.3 = 157.388 J At Point o height of the ball is by = R=2.1m final total energy TEJE PEH + kEf+RKEJ TEf= mghf 12m 0 + 12 20 TEF= (2-2x 9,8x2-1) + å muset za miny ball

TEJ= 45:276 + most smur TEG= 45.276+70 mux= 45.276 +1.540} from the conservation of 45.276 +1:54 V1 = 157.388 157.388-45.276 = 1120112 energy TE;=TEX 1.540 - of= 112:112 1.54 = 72.8 velocity of لها ac from the Uf = 8.532 figure Fr=mac Normal force e FN mu FN = 2.2 x 72.8 2.1 R Normal force on the ball FN=76.27.