Projectile Motion Lesson 1 - Horizontal Projectile Motion
Projectile motion occurs when one object has both horizontal motion and vertical
motion at the same time. The first type of projectile motion is called horizontal
projectile motion. This occurs when an object is shot straight out, not up at an angle,
and falls down until it hits the ground.
The main thing to remember when working with projectile motion, is that even though
the object has both horizontal and vertical velocity, those two components are to be
handled separately. When an object is shot horizontally from the same height, but with
different force, it will always take the same amount of time to land. The bigger the
initial force, the farther the object will travel, but they will all land at the same
time.
The reason is because the vertical motion, as stated earlier, is not dependant on
the horizontal. An object will always fall with an acceleration of -9.8m/s2. So, as
you can see, it takes the same amount of time to hit the ground, the variance is how far
it lands from the initial firing point.
So, as a general rule, the first thing you need to do in solving a projectile motion
problem is to find out the time in flight. Once that is found every thing else is
easy. Let's do an example.
A cannon shoots a cannon ball horizontally from a 24m high cliff, which lands 100m
from the base of the cliff.
a. With what velocity was the projectile fired?
b. What is the projectiles velocity as it strikes the ground?
In horizontal projectile motion, since the object is shot horizontally (hence the
name), its initial vertical velocity starts at 0m/s. Remember, we are going to start
with time. Since we know the height of the cliff and we know the vertical acceleration
of the cannon ball, we'll use the d = vot + 0.5*a*t2 formula.
-24m = 0m/s * t + 0.5 * -9.8m/s2 * t2
-24m = -4.9m/s2 * t2
4.898s2 = t2
2.213s = t
Now, according to Newton (about whom we will talk in a later lesson), an object will
travel at a constant velocity unless acted on by an outside force. Since there is no
outside force acting upon the cannon ball as it travels horizontally, we can assume that
it is moving at a constant velocity.
Wait!! What about gravity? Remember, gravity affects vertical movement, and since
vertical and horizontal are separate entities, it has no affect on the ball's horizontal
movement. So it is simple to calculate the objects initial velocity.
100m / 2.213s = vo
45.188m/s = vo, which is the answer to part a
For part b, since we know the acceleration, the initial velocity, and the time, we
can figure out vf easily.
-9.8m/s2 = (vf - 0m/s) / 2.213s
-21.687m/s = vf
Gee, that was easy. Wait, we're not done yet. At the beginning of the problem the
ball only had horizontal velocity, but now it has both, so we have to add them together.
If you don't remember how to do this check out this earlier
lesson. We just figured out the final vertical velocity, and since the horizontal
velocity is constant, we can just do some simple calculations and figure out the final
impact velocity.
sqrt ((45.1888m/s)2 + (-21.687m/s)2) = 50.123m/s
tan^-1 (-21.687m/s / 45.188m/s) = -25.638°
So the impact velocity is 50.123m/s at -25.638°.
The following lesson will delve into angled projectile motion.
