Motion with Constant Acceleration in Two and Three Dimensions
We have already seen that motion in more than one dimension that undergoes constant acceleration is given by the vector equation:
where a , v 0 and x 0 are constant vectors denoting the acceleration, intitial velocity, and initial position, respectively. Our next task will be to analyze special cases of this equation that describe important examples of two- and three-dimensional motion with constant acceleration: mainly, we will study projectile motion.
Simply stated, projectile motion is just the motion of an object near the earth's surface which experiences acceleration only due to the earth's gravitational pull. In the section on one-dimensional motion with constant acceleration, we learned that this acceleration is given by g = 9.8 m/s2 . Using a three-dimensional coordinate system, with the z -axis pointing upwards to the sky, the corresponding acceleration vector becomes a = (0, 0, - g) . This turns out to be the only piece of information we need to write down the general vector equation for projectile motion.
As an example, consider a creature fired out of a canon with speed v at an angle θ from the earth's surface. How far away will the creature be when it falls back down to earth?
The y -equation is pretty much useless. If we break this up into x - and z -components we get:
|z(t)||=||v sinθt - gt 2|
The next step is to find that time at which the creature will hit the ground. Setting z(t) = 0 and solving for t we find that the time at which the creature will hit the ground is t f = . Finally, we need to plug this time into the equation for the x -position, to see how far the creature has traveled horizontally in this time.
Using the trig identity sin(2θ) = 2 sinθcosθ we find that when the creature hits the ground its distance from the canon will be: