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viva [34]
2 years ago
15

In major league baseball, the pitcher's mound is 60 feet from the batter.If a pitcher throws a 89 mph fastball, how much time el

apses from when the ball leaves the pitcher's hand until the ball reaches the batter?
Physics
1 answer:
lianna [129]2 years ago
8 0
<span>Hitting can be broken down into three segments; SEE, REACT, SWING. You watch the ball in the pitcher's hand during the windup and you watch the ball leave the pitcher's hand when it is thrown and you watch the spin of the ball as it comes towards the plate. That is SEE. You determine what the pitch is (fastball, curveball, etc.), you determine where the ball is going to go and you determine whether it is headed towards an area that you think you can get good wood on it. That is REACT. You swing if you like the pitch or don't swing if you don't like the pitch. That is SWING. It doesn't make any difference whether you are playing with a tennis ball or a golf ball or a baseball. The theory and mechanics are the same. So, to answer your question I would say the more you practice the better you will be, regardless of the type of ball you use to practice with.</span>
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Answer:

v_{f} = 115.95 m / s

Explanation:

This is an exercise of a variable mass system, let's form a system formed by the masses of the rocket, the mass of the engines and the masses of the injected gases, in this case the system has a constant mass and can be solved using the conservation the amount of movement. Which can be described by the expressions

        Thrust = v_{e}  \frac{dM}{dt}

        v_{f}-v₀ = v_{e} ln ( \frac{M_{o} }{M_{f}} )

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the initial mass is the mass of the engines plus the mass of the fuel plus the kill of the rocket, let's work the system in SI units

       M₀ = 25.5 +12.7 + 54.5 = 92.7 g = 0.0927 kg

     

The final mass is the mass of the engines + the mass of the rocket

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       t = 1.90 s

Let's start by calculating the velocity of the gases relative to the rocket, where we assume that the rate of consumption is linear

          thrust = v_{e} \frac{M_{f} - M_{o}  }{t_{f} - t_{o}  }

          v_{e} = thrust  \frac{\Delta t}{\Delta M}

          v_{e} = 5.26 \frac{1.90}{0.080 -0.0927}

          v_{e} = - 786.93 m / s

the negative sign indicates that the direction of the gases is opposite to the direction of the rocket

now we look for the final speed of the rocket, which as part of rest its initial speed is zero

            v_{f}-0 = v_{e} ln ( \frac{M_{o} }{M_{f} } )

we calculate

            v_{f} = 786.93 ln (0.0927 / 0.080)

            v_{f} = 115.95 m / s

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