here is the problem: A pitcher throws a baseball from the mound to homeplate, a distance of 18.5m, in .411s. What was the average speed of that ball in miles/h? If the catcher allowed his mitt to recoil backward 7.5cm while stopping the ball, what was the acceleration of the ball while it was being brought to rest? Okay, the first part of the question I got, it was 101 mi/h, however the 2nd part has an answer of -1.4x10^4 m/s^2 for acceleration and I just cant seem to figure out how the number got that big...
hmm seems like the ball would be slowing down not speeding up if you assume that it was [tex] Vo=45 m/s (101mi/hr) Vf=0 x=.075m [/tex] [tex]x=.5(Vo + Vf)[/tex] [tex].075=.5(45)t[/tex] then t = .003 seconds [tex]V=Vo + at[/tex] [tex]0=45 + a*.003[/tex] [tex]a= aprox -13500 m/s^2 [/tex]
one question i dont c how the ball can decelarate at 13500 miles per hour, onless the pitcher is an alien i dont c that becoming posible.
The ball has a large initial speed and it has been slown down in a very small time interval. Thats why the ratio of the initial speed to this time interval is quite a large quantity (mathematically). Think of it like car moving at high speed brought to rest by a decelerating force. The magnitude of the deceleration produced is proportional to the initial velocity of the car and inversely proportional to the time taken for the car to come to rest. The smaller the time taken to do so and the larger the initial velocity of the car, the greater is the deceleration.
i see what you are saying but, if someone throws a ball at lets say 100 miles per hour calculate that in seconds will be like nothing. So how can it decelarate at 13500 miles per second.
if you are moving at a constant speed your deceleration is dependent upon how fast you slow from this speed since the time is so slow the deceleration must be very fast.... or the catcher is going to be forced to catch it over a longer time span