2.8 Game 2302-0380 motion -- variable velocity and acceleration  (Page 14/17)

Horizontal and vertical parameters

Following that, the code in Listing 4 defines horizontal parameters and vertical parameters that will be used to compute the horizontal and verticalcomponents of motion respectively.

Horizontal acceleration

The horizontal acceleration is set to zero. In the absence of air resistance, there is nothing to cause the horizontal component of the arrow to speed up orslow down until it stops suddenly when it hits the ground.

Vertical acceleration

The vertical acceleration is set to the acceleration of gravity at the surface of the earth, which is thesame value used for the previous exercises in this module.

Decompose the initial velocity

The cosine and sine functions are used to decompose the initial velocity into horizontal and vertical components of velocity.

Initial horizontal and vertical positions

Finally, the horizontal position when the arrow is released is set to 0 and the vertical position is set to 6 feet, approximately the height of the releasepoint for an archer that is six feet tall.

Working variables

After defining the horizontal and vertical parameters, the code in Listing 4 declares working variables for time, horizontal position (x), and verticalposition (y).

A while loop

A while loop is used to iterate for as long as the arrow is above the ground.

During each iteration, the equation given above is evaluated twice, once to determine the height of the arrow and once to determine the horizontal positionof the arrow at the current time.

Time starts at 0, and increments by +0.25 seconds during each iteration.

The document.write method is called inside the while loop to display the results shown in Figure 6 .

Acceleration of gravity exercise #3

A quadratic equation

Now let's turn things around and approach the problem from a different viewpoint. If we subtract d from both sides of the motion equation given above , replace a by g to avoid confusion later, and then rearrange the terms, we get the following equation :

0.5*g*t^2+v0*t-d = 0

where

• d is distance in units of distance
• v0 is the initial velocity in units of distance/time
• t is time in units of time
• g is acceleration in units of distance/time^2

In this form, you may recognize this as a standard quadratic equation which is often written as

a*t^2 + b*t + c = 0

(Now you know why I replaced the acceleration term a by the acceleration term g earlier. I need to use the symbol a in the standard quadratic equation.)

Standard solution for a quadratic equation

Given values for a, b, and c, you should know that the solution for determining the values for t is to find the roots of the equation.

There are two roots (hence two values for t). You should also know that the two roots can be found by evaluating the quadratic formula in two forms.

t1 = (-b+Math.sqrt(b*b-4*a*c))/(2*a);

t2 = (-b-Math.sqrt(b*b-4*a*c))/(2*a);

Using the solution

Relating the coefficients in the standard motion equation to the coefficients in the standard quadratic equation gives us:

• a = 0.5*g
• b = v0
• c = -d

The scenario

Let's use the scenario posed in the first exercise in this module. In this scenario, an archer that is six feet tall shoots an arrow directly upward with a velocity of 100 feet per second. Assumethat the arrow is at a height of 6 feet when it leaves the bow. Also ignore the effects of air resistance.