Projectile Motion
Up to this point, we've covered motion in one direction (side-to-side) and motion in multiple directions. Now, it might be useful to talk about projectile motion.
First, we should define a projectile. A projectile is defined to be an object moving only under the influence of gravity, where gravity is constant in both magnitude and direction. Projectiles also follow parabolic paths, when not under the influence of air resistance.
Some examples of projectiles are: different types of sports balls being thrown, hit, or kicked into the air; bullets and mortar shells fired from guns; and so on. Try to think of your own below.
Some examples of things that are not projectiles include planes and satellites. Why are neither of these things projectiles? Try to explain why in the space below.
Let's look at a typical path of a projectile:
This projectile is launched at some angle with an initial speed . Its path should resemble a parabola (although the limits of the drawing software in Microsoft Word are clearly on display).
In order to describe the characteristics of this path mathematically, we need to adapt some familiar formulas. You may recall from our discussions on straight-line motion:
We can adapt these to our situation by looking more closely at the initial velocity vector:
So, here we have the x and y components of the object's initial velocity. We can use these to modify our straight-line formulas to use in projectile motion, which result in the following four useful formulas:
The first row tells us where the object is along its flight path, in terms of its range (horizontal distance covered) and its height (vertical position). The second row tells us information about its velocity components. In both vertical cases, the minus sign is in front of because gravity points downward, in the negative direction.You may notice that there is a time and an acceleration term in the vertical velocity formula, but not in the horizontal velocity formula. Why is this so? (Hint: what forces are acting on a projectile?) What does this imply about the magnitude and direction of each of these velocity components?
In general, we should rely on these formulas to solve the problems we are given. However, some unique tricks may appear, in special circumstances. For instance, as we discussed, if a projectile starts and finishes at the same vertical position, then we have a parabola that can easily be split into two halves; the time of flight at the projectile's maximum height, , is half of the total time of flight. In addition, it is useful to note that when an object reaches its maximum height, its vertical component of velocity is zero. (Check this against what you wrote down in the previous blank section- are they compatible?) We can often use this knowledge to solve some questions, and use it to inform some of our thinking for others.
ex1:
A Duke basketball player attempts to drive in for a layup, but is blocked by a UNC player. Undeterred, the Duke player circles outside until she is forced nearly to the half-court line, where she heaves it before the shot clock gives out. She gives the basketball a velocity of , directed at an angle of . The difference in height between her hands and the rim of the basketball hoop is 1 m. If she is standing precisely 13.7 m horizontally away from the hoop, does he make the shot?
Solution:
Let's start by drawing a diagram.
So, now we've labelled all the information that we have been given, labelling our starting height as our "zero." The question is, does she make it or not? For this, we will need to solve for the horizontal distance this projectile will go. One method of doing this is by taking several steps and combining information: 1) find the time taken to reach the maximum height; 2) find the maximum height itself; 3) find the time taken to fall from the maximum height to the final height. A second method will be discussed after the solution.
We solve for the time taken to reach the maximum height by realizing that at the maximum height, its velocity is zero. So:
Then, we find the maximum height:
And now, knowing that it only falls 0.28 m- because the rim is 1 m higher than the hands that shot the ball- we can find the time taken to fall back down. However, starting at this position, the angle in the equation becomes , as the velocity at the top is directed entirely to the right:
The negative sign in front of the 0.28 is there because the ball has a negative displacement between the maximum height and its final position.
Now, to solve this, we need the quadratic formula. Simplify this:
Then:
For our purposes, we take the positive solution. Mathematically, the negative solution
corresponds to the other side of the parabola from the equation, meaning 0.24 s had elapsed from the time it went from 1 m to 1.28 m. (As a challenge, check this for yourself!) The total time of flight is the sum of the two pieces:
Now, we can use this to solve for the horizontal range!
She misses… badly!
Try the other method: we can put together one equation that captures the dynamics of the problem, and solve that one equation. Start with:
We know that the hoop is above the player’s hands. With being her hands, and up being positive:
Solve this:
Here, both values are positive. What do we do with this? Well, let’s think: if we were to trace the path of the ball, like in the diagram, how many times does the ball get to ? Twice! The first value corresponds to when the ball passes through this height, on the way toward the maximum height. The other solution corresponds to the other end of the parabola! Now that we have the total time of flight, we can use one of the other equations:
As with the first solution, we show that she misses… badly!
One more example, where things are symmetric:
ex2:
A kicker on a football team kicks the ball off, giving it a speed of , directed at an angle of . The receiver, realizing the ball is bound to head out of bounds, moves out of the way and lets it hit the ground untouched. Find the horizontal distance the ball covered.
So, we can do this one of two ways; we can do it using the method outlined in the previous example, or- because this is a parabola, with the starting and finishing vertical positions at zero- we can use the quadratic formula implied by the vertical position equation. I'll use both.
We solve for the time taken to reach the maximum height by realizing that at the maximum height, its velocity is zero. So:
Unlike the last example problem, this problem is truly symmetric- the time taken to reach the maximum height should equal the time to fall from that height. Thus, we can find the total time of flight by multiplying this time by 2:
Now, we can use this to solve for the horizontal range!
Now, we attempt the other method:
Method 2:
We write an equation to capture everything into a one-step equation, and solve. Start with:
Because it's a symmetrical parabola, and the starting and finishing points are both at . So:
From here, you can either use:
Set the parentheses equal to zero, which gives:
Or, if you prefer the quadratic formula:
or
Zero makes sense, in that at , the ball is at on a graph of the ball’s vertical
position.We, however, are interested in the horizontal displacement of the ball, so we take the nonzero solution.
Plug it in to find the range:
Below are some practice problems for you to try:
1. Johan Santana of the New York Mets can throw a fastball at 40 m/s. If he stands 18.44 m away from the batter and throws his pitches horizontally:
a. How long does it take for the ball to reach the catcher? b. In that period of time, how far does the ball vertically drop?
2. A velociraptor tears a piece of its prey off, but forgets to keep its jaw clenched when it does. As a result, it winds up being set into motion through the air. If the distance from the ground to the velociraptor's mouth is about 2 m, and it is launched from its mouth at a speed of 5 m/s, and at an angle of how far must the velociraptor now walk to eat this morsel?
3. In a part of a Mythbusters episode, Adam made sure his gun rig was firing level by positioning a board 20 ft (6.1 m) away from the barrel. He said that “in this time frame, the bullet would not drop.” Is he right, or mostly right? Answer this question by doing a calculation of how far a bullet would drop in that distance, assuming that the muzzle velocity of the gun was around 300 m/s.
Challenge Problem:
4. Mike Piazza is one of the best-hitting catchers in baseball history. In one instance, he hit a home run that landed about 120 m from home plate, and wound up approximately 10 m above where it left his bat at home plate. Supposing that the ball left his bat at an angle of
, find the speed at which the ball left his bat.
