A 5 1/8 oz baseball traveling at 80 mph rebounds off of a bat with a speed of 16

Question

A 5 1/8 oz baseball traveling at 80 mph rebounds off of a bat with a speed of 160 mph. The ball is in contact with the bat for roughly 10^-3 seconds. The incoming velocity of the ball is horizontal and the outgoing trajectory forms an angle of 31 degrees with respect to the incoming trajectory. What is the impulse provided to the baseball by the bat? Determine the average force exerted by the bat on the ball? Determine how much the angle would change ( with respect to 31 degrees) if we were to neglect the effects of the force of gravity on the ball? FBD WOULD BE PERFECT SO I COULD UNDERSTAND THE ANSWERS MORE CLEARLY A 5 1/8 oz baseball traveling at 80 mph rebounds off of a bat with a speed of 160 mph. The ball is in contact with the bat for roughly 10^-3 seconds. The incoming velocity of the ball is horizontal and the outgoing trajectory forms an angle of 31 degrees with respect to the incoming trajectory. What is the impulse provided to the baseball by the bat? Determine the average force exerted by the bat on the ball? Determine how much the angle would change ( with respect to 31 degrees) if we were to neglect the effects of the force of gravity on the ball? FBD WOULD BE PERFECT SO I COULD UNDERSTAND THE ANSWERS MORE CLEARLY

Explanation / Answer

w = 5.125oz / (1lb/16oz) = 0.3203125 lb
m = w/g = 0.0099554 slug

v1 = -80 mph = -177.33 ft/s
v2 = 160 mph = 234.67 ft/s

v1x *m + integral(Fx)dt = v2x*m

-177.3*0.0099554 + integral(Fx)dt = 234.67*cos(31)*0.0099554
Ix = integral(Fx)dt = 3.768 lb s

v1y *m + integral(Fy)dt = v2y*m
Iy = integral(Fy)dt = 234.67*sin(31)*0.0099554
Iy = 1.2032 lb s

I = sqrt(Ix^2 + Iy^2)
I = 3.9955 lb s

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