Path and Range of a Baseball
Baseball is like church, many attend but few understand
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
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,
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 value x is the range of the ball when it is hit with x velocity vx0 and y velocity vy0.