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Thread: Applicatin - Horizontal Distance

  1. #1
    Nov 2007

    Applicatin - Horizontal Distance

    Horizontal distance traveled by a projectile fired from the ground can be modeled by $\displaystyle d = \frac {v_0^2\sin{2\theta}}{g}$, where $\displaystyle v_0$ is the initial velocity, $\displaystyle \theta$ is the launch angle in degrees, and "g" is acceleration due to gravity. (This model neglects air resistance and assumes a flat surface.)

    a) A baseball is hit at an initial height of 5.5 feet with initial velocity of 95 ft/s at 42 degrees to the ground. Will the ball clear a 5 ft fence 310 feet from home plate?

    I used a calculator to solve this one. I plugged in $\displaystyle y = \frac {95^2\sin{2x}}{9.8} +5.5$ and find that when "x" is 42, the value of "y" is 921.3735. Since the "y" value is horizontal distance, the distance from home plate is 921.3735. My question is how do I know that the ball cleared the 5 ft fence 310 ft away from home plate, since the "y" value is only the horizontal distance and not vertical distance?

    b) What angle will maximize the horizontal distance traveled and why?

    I find out that the angle that will maximize horizontal distance is 45 degrees, which is true in physics from what I've learned. At 45 degrees the baseballl will travel 926.42 feet. My question is why is 45 the angle that maximizes horizontal distance?

    c) Will the ball clear the fence if it at the angle from part (b) above?

    Just like in part (a), how do I know that the ball cleared the 5 ft fence 310 ft away from home plate? I know it traveled 926.42 feet horizontally, but I don't know if it cleared the 5 foot fence.
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  2. #2
    May 2009

    Solution to baseball clearing fence physics problem

    I am going to start with the equation $\displaystyle x= x_{0}+v_{0}t+\frac{1}{2}at^2$ where x represents distance, a is acceleration and t is time and $\displaystyle v_{0}$ is initial velocity

    So the total horizontal distance the ball will travel is $\displaystyle x=0+95\cos(42)t+\frac{1}{2}(0)t^2$ as there is no air resistance so there is no acceleration in the horizontal direction and 95cos(42) is initial velocity in the horizontal direction. So x=95cos(42)t and we need to find what t is by finding out how long the ball will be in the air.

    So for the vertical distance and time, we want the ball to end up on the ground, so height = 0... so using the same distance equation

    $\displaystyle 0=5.5+95\sin(42)t+\frac{1}{2}(-9.8)t^2$ and solving this we find that t=13.059 (we ignore the negative value for t, it arises because this function is a parabola)

    So now that we have t=13.059, x=95cos(42)*13.059=921.942

    Using a bit of physical intuition, if a ball travels over 900 feet, then it must be higher than 5 feet off the ground when it has travelled 310 feet (under any type of normal baseball conditions anyway), still to prove it...

    We know need to find how long it takes the ball to reach 310 feet, so using the horizontal equation, $\displaystyle 310=0+95\cos(42)t+\frac{1}{2}(0)t^2$ and t=4.391

    And so now we need to find how high the ball is at time t=4.391
    $\displaystyle y=5.5+95\sin(42)*4.391+\frac{1}{2}(-9.8)*4.391^2$ and we get y=190, so when the ball is 310 feet from home plate, it is 190 feet in the air, so it clears the fence and the hitter is probably on steroids...

    45 degrees maximizes horizontal distance because at 45 degrees, the initial velocities in both the x and y directions are equal. This leads to maximizing horizontal distance
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