Question

A 5 oz. baseball has an initial horizontal velocity of 110 ft/sec at the instant it is struck by a
bat. The ball is in contact with the bat for 0.15 seconds and leaves the bat with a velocity of 140
ft/sec at 30 degrees with respect to the horizontal. Find the magnitude of the average force
applied to the ball.

Answer 1

F_x = m (vcos30 - v₀) / t

= (5)(2.8349*10^-2)(140cos30-(-110))(0.3048) / 0.15

= 66.604 N

F_y = mvsin30/t

= (5)(2.8349*10^-2)(140sin30)(0.3048) / 0.15

= 20.162 N

F = sqrt(F_x² + F_y²) = 69.589 N

= 15.64 lb

Answer 2

Fx =m*(140cos30-110)/0.15 =374.785

Fy =m*140sin30/0.15 =2333.333

F =[Fx^2 +Fy^2]^0.5 =2363.241 lb_f

Answer 3

m = (5/16)/32 = 9.76 * 10E-3 slug
Vx = 110 + 140 cos 30 = 231 m/s change in horizontal velocity
Vy = 140 sin 30 = 70 m/s change in vertical velocity
V = (231^2 + 70^2)^1/2 = 237 m/s
F t = m V impulse = change in momentum
F = 9.76 * 10E-3 * 237 / .15 = 1.54
Perhaps t = .015????