Path and Range of a Baseball
Baseball is like church, many attend but few understand
-Wes Westrum, San Francisco Giants coach (1962)
Editor's note: for sake of simplicity, we will assume that the baseball
will hit the bat in the "ideal"
place.
Check out
an applet modeling the path and range of a baseball in an ideal
situation
II. How far will the ball go?
Dr. Dell says:
Instead of just talking about it, lets find out the actual numerical distance traveled by a baseball.
The ball is hit with vx0, velocity in the x (horizontal) direction, and vy0, velocity in the y (vertical) direction.
We know of two equations we can use:
The initial velocity in the x direction equals the horizontal distance divided by the time
y = vy0 t - .5 a t2
The vertical
distance (height) equals the inital velocity in the y direction times the
time minus one-half the acceleration times the square of the time
Both equations are derived from the definition of position, velocity, and acceleration
Using the first equation, vx0=x/t, we can rearrange the terms by multiplying both sides by t (time) and dividing both sides by vx0 (horizontal velocity) to get a new equation in terms of t (time):
Time equals the horizontal distance divided by the horizontal velocity
We can now substitute this equation for t into our other equation,
y
= vy0 t - .5 a t2, to get yet another equation:
The vertical distance (height) equals the initial vertical velocity times the horizontal distance divided by the initial horizontal velocity MINUS one-half the gravitational constant times the square of the quotient of the horizontal distance divided by the horizontal velocity
Where did the g (gravitational constant) come from? Well, since the ball is moving through the air, gravity will be the force causing it to fall back down. Thus, in y = vy0 t - .5 a t2, the acceleration a towards the earth will be equal to gravitational constant g, which is about 9.8 meters per second per second (yes, two "per seconds". That simply means that every second, the velocity of the object increases by about 9.8 meters per second).
Don't be scared by the complexity of this equation! The trick is that when the ball hits the ground, y will be 0 (as well when the ball is originally hit). So the new equation will be:
add .5 g (x / vx0) to both sides:
vy0 x / vx0 = .5 g (x / vx0)2
Divide both sides by (x / vx0):
vy0 = .5 g (x / vx0)
Multiply both sides by vx0:
vx0vy0 = .5 g x
And divide both sides by .5 g to get:
x = 2 vx0 vy0 / g
the horizontal
distance equals 2 times the initial horizontal velocity times the initial
vertical velocity divided by gravity
The value x is the range of the ball when it is hit with x velocity vx0 and y velocity vy0.
In the following JavaScript test, we will use velocity and angle. Don't worry, we don't have to derive a new equation. Through vectors we know that vx0=velocity*cos(angle) and vy0=velocity*sin(angle); take our word for it.
