Baseball problem Impulse-Momentum

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A baseball weighing 4 ounces is thrown at 60 ft/s and then hit at a 40-degree angle with a velocity of 120 ft/s. The average force applied during the impact is calculated using the impulse-momentum principle, resulting in a force of 66.5 pounds. This force seems low compared to typical baseball impacts, which occur at much higher speeds and shorter contact times. In professional baseball, pitch speeds and impact durations lead to forces that can exceed 10,000 pounds. The calculations and context provided help clarify the reasonable nature of the calculated force for this scenario.
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A baseball (4oz .00781 slugs) is thrown at 60 ft/s. After it strikes the bat it is hit at 40 degrees to the initial trajectory (towards third base) and then has a velocity of 120 ft/s. What is the magnitude of the average force applied? T=0.02 sec.



2. F delta T = m delta V



3. Delta Vx= 120 cos 40 - (-60) = 151.9
Delta Vy= 120 sin 40 = 77.13

DeltaV=Square root(1519.9^2+77.1362^2)=170.4

F= .00781*170.4/.02sec =66.5 ft lb

This does not seem right to me. Any help would be appreciated.
 
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A slug is the mass. 4oz/16 =.25/32ft/s^12=0.00781 slugs

Are you sure that is right. The force seems low to me.
 
Anarion said:
A slug is the mass. 4oz/16 =.25/32ft/s^12=0.00781 slugs

Are you sure that is right. The force seems low to me.
Ahh, America! The force is in pounds, not ft-lbs. The impact force does seem low at first, but when you consider the ball is thrown at only 60 ft/sec (about 40 mph, a little league pitch , and less than that of a Wakefield 60 mph knuckle ball, coupled with the fact that the impulse time of 0.02 seconds is 'large', and the ball is not hit squarely back toward the pitcher, then the 66.5 pounds force is reasonable. In a major league baseball game, the speed of the pitch is much faster, and most of all, the impact time of the ball in contact with the bat is much, much smaller, like .001 seconds or less. Thus the impact force could be about 100 times larger, or even more (up to 10,000 pounds in some cases).
 
Thanks for your help!
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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