How Much Does Gravity Affect a Baseball's Trajectory Over 18.4 Meters?

AI Thread Summary
The discussion focuses on calculating the drop of a baseball due to gravity over a distance of 18.4 meters when thrown horizontally at 76.4 mi/h. The time taken for the ball to reach home plate is determined to be approximately 0.542 seconds. Using the vertical motion equation, the drop is calculated as y = 0.5 * a * t^2, resulting in a drop of about 1.44 meters. The analysis assumes no air resistance and a perfect pitch, simplifying the vertical component to just the effects of gravity. This calculation highlights the significant impact of gravity on a baseball's trajectory over the given distance.
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1. While trying out for the position of pitcher on your high school baseball team, you throw a fastball at 76.4 mi/h toward home plate, which is 18.4 m away. How far does the ball drop due to effects of gravity by the time it reaches home plate? (Ignore any effects due to air resistance and assume you throw the ball horizontally.)
2. 1. x=1/2(v0+vf)t; 2. x=v0t+1/2at^2
3. I split up the values given into a horizontal and vertical table. Since neither vertical or horizontal has time, I solved horizontal for time and got .542s. I then took that and plugged into the 2nd equation listed, and got 19.8506, which was incorrect. My vertical and horizontal table looks like the following:

Vertical- a=9.8 m/s/s, h=?, t=
Horizontal- x=18.4 m, a=0, v0=33.975 m/s, vf=33.975

Any help asap would be great thank you!
 
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Your approach looks good. Let's say everything is correct for the horizontal component...
In the vertical component you have
y = y_0 + v_y t + \frac{1}{2} a t^2
You are really just interested in how much the ball is dropping so let's say y_0 = 0 .
We'll also assume this is a perfect pitch so v_y = 0 .

So we end up with y = \frac{1}{2} a t^2 = 0.5(-9.8)(0.542^2) = -1.44 m
 
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