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Projectile Motion What happens when we give an object both an initial horizontal and vertical speed? The result is pictured below.  This is still a parabolic path. In fact, the last half of the path is identical to the path taken by ball that rolls off of the table.  There are a couple of things to note about this picture. First, notice how the two sides are mirror images of each other. Second, at the very top of the trajectory, notice how the dashed line is completely horizontal. Remember when an object is thrown straight into the air, the vertical velocity is zero at the very top of its path? This is still true here. It does not matter if an object is given a horizontal velocity when it is shot up into the air. That horizontal velocity has no affect on the vertical velocity and vice versa. So, again we are dealing with an object doing two things simultaneously: it is traveling horizontally and vertically at the same time. Therefore, the object does not have any vertical velocity the moment it hits the top of its trajectory, just as an object that is thrown straight into the air with no horizontal velocity. Let's examine this further to understand these things. The first thing we need to address is the moment the ball is shot out of the cannon. It is given both a horizontal velocity and a vertical velocity. They are called the components of the ball's velocity out of the cannon.  The picture above shows the velocity vector, in red, and the two components Vx and Vy, in black. Obviously the Vx represents the horizontal part of the velocity, and Vy represents the vertical part. We have to use trigonometry to find these components. ( If you are unfamiliar with trigonometry click here for a brief synopsis). The picture below shows the trig equations to breakdown the velocity vector.  The Vx component requires you to take the velocity and multiply it by the cosine of the projectile's angle. Likewise, the Vy component requires the sine function. For example, if the object's velocity is 100 m/s with an angle of 45 degrees, the Vx and Vy components are found by,  This results in Vx = 70.7 m/s and Vy = 70.7 m/s. They happen to be the same because of the angle. I will address that more later. Just know that they will not always be the same. Let's address each compenent and their consequences. Horizontal Component Since the horizontal component is not affected by gravity, and we are not considering air resistance, it stays the same value throughout the flight of the projectile.  Because it stays the same, the ball travels the same horizontal distance every second. Vertical Component The vertical component is affected by gravity. As soon as the ball leaves the cannon the vertical speed begins to decrease until it reaches zero. After that instant, the vertical speed begins to increase.  This is exactly the same thing depicted in the picture found in chapter one.  Notice how in both cases, the vertical speed starts out at a certain length and decreases until it reaches zero. Then, it begins to increase in the opposite direction. The only difference between the two pictures is the first ball travels a horizontal distance while the vertical speed changes. This means we can still use the free fall equations with projectile motion.  Before we get to that, let's see both the vertical and horizontal components in the picture, using the vector addition rules. The red arrow is the velocity, which is the resultant, of the ball throughout the path.  Notice how the red arrow is pointing completely in the horizontal direction at the top of the trajectory. Again, that is because, at that moment the velocity of the object consists solely of the horizontal component. After seeing how the two components work, it should be no surprise that the two halves of the trajectory are mirror images of themselves. We can now work through an example to see how this all fits together. A cannon is shot with a velocity of 150 m/s 30.0 degrees above the horiztonal (I use AH for above the horizontal and BH for below Horizontal). Find the maximum height attained by the cannon, and how far it travels from the firing point. The first thing we need to do is find how long it takes to reach the maximum height using the equation,  We can use this because we know the vertical velocity at the top, zero. Remember, this equation only deals with vertical velocities, which means Vo is not 150 m/s! The Vo is the initial Vy given by, Therefore, the velocity equation becomes, Solving for t gives 7.7 seconds. Now all we have to do is plug this time into the vertical position equation, Plugging in the numbers gives, Which yields a maximum height of 287 meters. Since the left side of the path and right side of the path are mirror images of themselves, it stands to reason that the time for the cannon to go from the top of its trajectory back to ground level would be the same as the time it took to get to the maximum height. In fact, that is exactly the case.  Therefore, the total time of the trip would be 7.7 + 7.7 seconds, which equals 15.4. Now that we know the total time, we can find the horizontal distance traveled. During the entire 15.4 seconds of the trip, the cannon was traveling horizontally with a speed equal to the horizontal component of the velocity given by, In order to find the distance we need to use the speed equation, Therefore, Solving for x gives a distance of 2000 meters.  There is another equation to use to find the maximum distance directly,  In this equation, Vo is the velocity of the cannon and not just Vy. Plugging in the numbers yields, The actual answer is equal to 1988 meters. But, using significant digits, the answer does equal the 2000 meters found above. (The time found above was rounded. If all of the digits were used, the horizontal distance would have been 1988 meters. It does not really matter in this case because when you consider the significant digits in this problem, you can only end up with 2 significant numbers in your answer. In both cases the number rounds to 2000). Distance Fallen There is a very cool fact associated with projectile motion. Consider the picture below.  The straight red line is the path the cannon would travel if there was no gravity pulling it to the ground. Remember from chapter one how far an object falls after one second? The answer is 4.9 meters. Guess how far the cannon is from the red line after one second? That's right, 4.9 meters! It turns out that even though an object can be shot at any angle and any direction, it will fall from its original line it would have taken exactly the same distance an object falls in normal free fall.  Complimentary Angles Another cool concept with projectile motion is the horizontal distance associated with complimentary angles, which are angles that add up to 90 degrees. Consider the picture below.  It turns out that any two complimentary pairs of angles will result in the same horizontal distance traveld during the flight even though they do attain different maximum heights. Notice, however, the compliment to 45 degrees is 45 degrees. Can you guess which angle yields the maximum horizontal distance? That's right, it is 45 degrees. No other angle is compliment to 45, so it is by itself. Also, it is the only angle where the Vx and Vy components are equal. (Remember the example before). The picture above is to actual scale, so you can see how the heights and horizontal distances compare for the given angles. |