# Baseball and 2D Kinematics

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 in to 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!

## Answers and Replies

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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 lets say $y_0 = 0$ .
We'll also assume this is a perfect pitch so $v_y = 0$ .

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