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# Projectile Motion

## When an object is thrown, gravity acts upon it and affects the path traveled.

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Projectile Motion

### Lesson Objectives

The student will:

• draw and interpret graphs involving two-dimensional projectile motion
• solve for the instantaneous velocity of a projectile
• predict a projectile’s range

### Vocabulary

Free fall
The condition of acceleration which is due only to gravity.  An object in free fall is not being held up, pushed, or pulled by anything except its own weight.  Though objects moving in air experience some force from air resistance, this is sometimes small enough that it can be ignored and the object is considered to be in free fall.
Projectile motion
Projectile motion is a form of motion where an object (called a projectile) is thrown near the earth's surface with some horizontal component to its velocity.  The projectile moves along a curved path under the action of gravity. The path followed by a projectile is called its trajectory. Projectile motion is motion in two directions. In the vertical direction, the motion is accelerated motion and in the horizontal direction, the motion is constant velocity motion.
Instantaneous velocity of a projectile
Instantaneous velocity is the velocity of an object at one instant during its motion. In the case of a projectile, the instantaneous velocity vector would be the resultant of a constant velocity horizontal motion and an accelerated velocity vertical motion.
Range
A projectile launched with specific initial conditions will travel a predictable horizontal displacement before striking the ground. This distance is referred to as the projectile's range.

### Equations

\begin{align*}y = \left(\frac{v_i+v_f}{2}\right)t\end{align*}

\begin{align*}x_f = (v \cos \theta)t + x_i\end{align*}

\begin{align*}y_f = \frac{1}{2} gt^2 + (v \sin \theta)t + y_i\end{align*}

\begin{align*}v_x = v \cos \theta\end{align*}

\begin{align*}v_y = v \sin \theta\end{align*}

### Introduction

“Independence of Motion along Each Dimension” looks at the special case of throwing an object in a purely horizontal direction.  Projectile motion is the general case of throwing an object in any direction, from sliding an object off a desk to kicking a soccer ball as shown in the Figure below.

http://demonstrations.wolfram.com/JumpingOverRowOfParkedCars/

#### Finding instantaneous velocity

Let's look back at the case of a dart fired out from a dart gun, using vector mechanics.  Consider the trajectory of the dart sometime between pulling the trigger and striking the ground. The diagram below is a representation of the instantaneous velocity components of the dart.

Credit: CK-12 Foundation

Author: CK-12 Foundation - Raymond Chou License: CC-BY-NC-SA 3.0[Figure1]

Using only one-dimensional mechanics, we found that in the y-direction, it takes 0.55 seconds for the dart to hit the floor from a height of 1.5 meters.  If the dart traveled a horizontal distance of 6.0 m in 0.55 seconds, then its horizontal component of velocity is therefore \begin{align*}v_x=\frac{6.0\text{m}}{0.55\text{s}}=10.9 \ \text{m/s}\end{align*}.

What is the vector velocity at time, \begin{align*}t = 0.25 \text{s}\end{align*} ?

The \begin{align*}x\end{align*} and \begin{align*}y\end{align*} velocity components represent the legs of the right triangle (see Figure above) and the hypotenuse represents the instantaneous resultant velocity of the projectile. Using the Pythagorean Theorem, \begin{align*}v_x{^2}+v_y{^2}=v^2\end{align*}, gives the magnitude of the instantaneous velocity of the projectile once the square root is taken.

The \begin{align*}x\end{align*} velocity is constant, so we know it is \begin{align*}10.9 \ \text{m/s}\end{align*} at all times. For the \begin{align*}y\end{align*} velocity, we know from one-dimensional motion that \begin{align*}v_f=at+v_i\end{align*}. The acceleration from gravity is constant, giving: \begin{align*}(-10 \text{m/s}^2) (0.25\text{s}) + 0\text{m/s} = -2.5 \ \text{m/s}\end{align*}

Now that both legs of the right triangle are known, we can apply the Pythagorean Theorem to solve for the instantaneous speed (the hypotenuse of the right triangle) at time, \begin{align*}t = 0.25 \ s\end{align*}.

\begin{align*}v = \sqrt{v_x^2 + v_y^2} = \sqrt{(10.9\text{m/s})^2+(2.5\text{m/s})^2}= 11.2 \ \text{m/s}\end{align*}

If all that was asked for in this problem was the instantaneous speed of the dart at 0.25 s we would be done. However, you may be asked for the dart’s instantaneous velocity at \begin{align*}t = 0.25 \ s\end{align*}. In that case, you need to determine the instantaneous direction of the dart as well (recall that velocity is a vector quantity, and as such, has a magnitude and a direction).

Finding the instantaneous direction of the velocity of the dart at \begin{align*}t = 0.25 \ s\end{align*}:

Using the tangent function is the most efficient method. The angle that we need to find is somewhat arbitrary; after all, other than the right angle, the triangle has two perfectly good angles to choose from. The angle usually preferred, however, is measured from \begin{align*}x-\end{align*}direction. Using the inverse tangent relationship which can be found on your calculator, we have:

\begin{align*}\tan \theta = \frac{2.5\text{m/s}}{10.9\text{m/s}} \ ; \ \tan^{-1} \theta = 12.7^{\circ}\end{align*}

If you are asked to represent the vector using the trigonometric definition of angle measurement, than either \begin{align*}-12.7^{\circ}\end{align*} or \begin{align*}347.3^{\circ}\end{align*} would do. The final vector result can be stated as \begin{align*}(11.2\text{m/s}, -12.7^{\circ})\end{align*}.

#### Components of projectile motion

In the problem that follows we imagine the projectile launched from the ground with an angular elevation between 0 and 90 degrees, with an initial \begin{align*}x-\end{align*}component of velocity of +30 m/s and an initial \begin{align*}y-\end{align*}component of velocity of +40 m/s.

Author: CK-12 Foundation - Christopher Auyeung License: CC-BY-NC-SA 3.0

Some typical questions that can be asked in such a situation are:

1. What is the time the projectile takes to reach the highest position above the ground?
2. What is the projectile’s highest position above the ground?
3. What is the velocity of the projectile at its highest position above the ground?
4. What is the range of the projectile?

1. In order to answer the first question, let’s consider what determines the amount of time the projectile remains airborne. A velocity of 30 m/s in the \begin{align*}x-\end{align*}direction, does not affect the time in the air; no more so than the gun or dart’s airborne time was affected by their horizontal motion. The vertical component of motion, however, must determine how much time the projectile spends airborne since only the vertical motion is subject to gravity. If we imagine just the vertical component of +40 m/s we can quickly estimate (if we make the approximation that -9.8 is close enough to -10) the amount of time it takes the projectile to reach its highest position (peak); it’s about 4 seconds, or if we need more accuracy; \begin{align*}\frac{40}{9.8} = 4.08 \ s = 4.1 \ s\end{align*}. Remember that the projectile loses 9.8 m/s of velocity every second it ascends. Thus,

1. Answer: \begin{align*}t = 4.1 \ s\end{align*}
2. Answer: Once again we can use \begin{align*}x=vt\end{align*} that is, \begin{align*}y=\frac{v_i+v_f}{2}t\end{align*}, as long as we understand that \begin{align*}v\end{align*} represents the average velocity for the ascent part of the projectile’s motion. Since the projectile began with an initial velocity in the \begin{align*}y-\end{align*}direction of 40 m/s and has a final velocity of 0 m/s, at the top of its motion in the \begin{align*}y-\end{align*}direction, its average velocity is \begin{align*}\frac{(40 + 0)}{2} = 20 \ m/s\end{align*}. Its highest position above the ground is therefore, \begin{align*}(20 \ m/s)(4.08) = 81.6 = 82 \ m\end{align*}.
3. Answer: It is often incorrectly thought that the velocity of the projectile at its peak position is zero. It must be remembered that until the projectile hits the ground, it is always moving horizontally. Though the vertical velocity is zero at the top of its motion, the horizontal velocity is still +30 m/s. Therefore, the velocity of the projectile at its highest position above the ground is +30 m/s (the + indicating the projectile is moving with a velocity of 30 m/s to the right).
4. Answer: The range of the projectile is determined by finding the total time the projectile is airborne and multiplying that time by its speed in the horizontal direction \begin{align*}(x=vt)\end{align*}. The projectile’s motion is symmetric in the absence of air friction so the total time for the trip is twice 4.08 s or 8.16 s. The range is, therefore, \begin{align*}(30)(8.16) = 244.8 \ m = 240 \ m\end{align*}.

Of course, instead of using a bit of physical reasoning and simple arithmetic, we could have used more sophisticated equations to answer all these questions, but why make things more complicated than they have to be?

Before leaving this problem, let’s alter the given information, in order to see another way the problem could have been stated.

A projectile is launched from the ground at an angle of 53.13 degrees with a speed of 50 m/s.

Find the answers to questions 1-4 above.

Our claim is that both problems are, in fact, identical and have the same answers. If we’re not given the horizontal and vertical components of the velocity, it is good policy to find them before trying to solve a problem of this sort.

The \begin{align*}x-\end{align*}component (horizontal) is: \begin{align*}v_x = v \cos \theta \rightarrow 50 \cos 53.13^{\circ} = 30 \ m/s\end{align*}

The \begin{align*}y-\end{align*}component (vertical) is: \begin{align*}v_{yi} = v \sin \theta \rightarrow 50 \sin 53.13^{\circ} = 39.999 \rightarrow 40 \ m/s\end{align*},

Since we see that the components are identical to the original problem, we can solve the problem the same way.

http://demonstrations.wolfram.com/ThrowingABaseballFromTheOutfieldToHomePlate/

An interesting aside: If at 4.1 s, (the time when the projectile has reached its maximum height) we pretend to erase the first half of the projectile’s motion and label the projectile at its peak position with a vector (rather an a component of a vector) pointing to the right with a speed of 30 m/s, the remaining path of the projectile would be similar to the bullet and dart of the previous section (3.3 “A Special case of Projectile Motion”) and the problem would be solved in the same manner as both bullet and dart problems. See Figure below.

Author: CK-12 Foundation - Christopher Auyeung License: CC-BY-NC-SA 3.0

Let’s take a trip!

The package remains under the plane, but people on ground see the package's descent as a parabola. Author: Image copyright Konstantin Yolshin, 2012; modified by CK-12 Foundation - Christopher Auyeung License: Used under license form Shutterstock.com Source: http://www.shutterstock.com

Imagine a small airplane flying at a constant elevation of 80 m over your school soccer field, with a velocity of 50 m/s, (about to 110 mph), eastward; see Figure above. Ignore air resistance.

1. What would the motion of the package look like to an observer on the ground?

Answer: An observer on the ground would see the package fall in a parabolic arc, as if it had been projected horizontally with a speed of 50 m/s; the same motion performed by the bullet and dart; (though the package would remain directly below the airplane).

2. What would the motion of the package look like to an observer on the plane?

Answer: The package would appear to fall straight down, since the package and the airplane both have the horizontal velocity of 50.0 m/s.

3. The plane releases the package when it flies over a target marked \begin{align*}A\end{align*} on the ground, will the package land on \begin{align*}A\end{align*}?

Answer: No, it will not, since, as the package falls, it is moving at the horizontal velocity 50 m/s.

4. How far from point \begin{align*}A\end{align*} does the package land?

If we know how much time the package spends traveling at 50.0 m/s after it is released, we can use \begin{align*}x=vt\end{align*} to solve the problem.

The time the package remains in the air is found using our one-dimensional equation:

\begin{align*}y_f = \frac{1}{2}at^2 + v_{iy}t + y_i\end{align*} where \begin{align*}a=g=-10 \ \text{m/s}^2 \ , \ v_{iy} = 0 \ \text{m/s}\end{align*} (at the instant the package is released it has no velocity in the \begin{align*}y-\end{align*}direction), \begin{align*}y_f=0\end{align*} and \begin{align*}y_i=100 \ m\end{align*}; therefore, \begin{align*}\frac{1}{2} (-10\text{m/s}^2) t^2 + 0 + 80\text{m}= 0\end{align*}. Solving for \begin{align*}t\end{align*}, we find \begin{align*}t = 4.0 \ \text{s}\end{align*}.

Thus  \begin{align*} x = (50.0\text{m/s})(4.0\text{s}) = 200 \text{m}\end{align*}. The package fell 200 m past target \begin{align*}A\end{align*}.

In order for the pilot to hit target \begin{align*}A\end{align*}, the package must be released 200m before reaching the target, or 4.0 seconds before reaching target \begin{align*}A\end{align*}. See Figure below

#### Deriving some general results from the projectile equations

From the problems that we discussed, it should be reasonably clear that the equations we derived for one-dimensional kinematics can be used, with caution, for 2-dimentional kinematics. The only difference is that care must be taken in correctly identifying velocity components. Recall that if velocity is stated as \begin{align*}V\end{align*} at an angle of \begin{align*}\theta\end{align*} then the initial \begin{align*}x\end{align*} and \begin{align*}y\end{align*} components of the velocity are, \begin{align*}v \cos \theta\end{align*} and \begin{align*}v \sin \theta\end{align*}, respectively. With this in mind, we rewrite our kinematic equations as follows:

1. \begin{align*}x-direction: x_f=R=(v \cos \theta)t+x_i\end{align*}

2. \begin{align*}y-direction: y_f=\frac{1}{2}gt^2+(v \sin \theta)t+y_i\end{align*}

The other equations are treated the same way, using the initial velocity in the \begin{align*}y-\end{align*}direction:

3. \begin{align*}v_{yf}=gt+v \sin \theta\end{align*}

4. \begin{align*}v_{yf}{^2}=(v \sin \theta)^2 + 2g \Delta y\end{align*}

5. \begin{align*}v_{ave}=\frac{v \sin \theta + v_{fy}}{2}\end{align*}

#### Special cases

What is the maximum range, \begin{align*}R\end{align*}, of a projectile if \begin{align*}y_i=y_f=0\end{align*}?

setting \begin{align*}y_i\end{align*} and \begin{align*}y_f=0\end{align*} in equation 2, and factoring out \begin{align*}t\end{align*}, we have:

\begin{align*}t\left(\frac{1}{2}gt+v \sin \theta \right)=0\end{align*}

There are two solutions:

\begin{align*}t=0\end{align*} and \begin{align*}t=\frac{-2v \sin \theta}{g}\end{align*}

The trivial condition is satisfied at launch \begin{align*}(t=0,y_i=0)\end{align*} and the nontrivial condition is satisfied at landing \begin{align*}\left(t=\frac{-2 \sin \theta}{g}, y_f=0 \right)\end{align*}.

If \begin{align*}t=\frac{-2v \sin \theta}{g}\end{align*} is substituted into equation \begin{align*}1:R=(v \cos \theta)t\end{align*}

We have: \begin{align*}R=(v \cos \theta)\left(\frac{-2v \sin \theta}{g}\right)=\frac{-2v^2 \sin \theta \cos \theta}{g}\end{align*}

Using the trigonometric identity \begin{align*}2 \sin \theta \cos \theta = \sin 2 \theta\end{align*}, we have:

\begin{align*}R=\frac{-v^2 \sin 2 \theta}{g}\end{align*}

We can extract a useful result from the range equation. Assuming that \begin{align*}g\end{align*} is constant, the range is a function of the \begin{align*}V\end{align*} and \begin{align*}\theta\end{align*}. Let’s consider the angle. Since the maximum value of the sine is 1.0 then whatever angle makes \begin{align*}\sin 2\theta=1\end{align*}, will also maximize the range. Since \begin{align*}\sin 90^{\circ}=1\end{align*}, \begin{align*}2 \theta = 90^{\circ}\end{align*} and the angle which produces the greatest range is \begin{align*}45^{\circ}\end{align*}. The range continually increases with increasing velocity so there is no “interesting” information we can get out of \begin{align*}V\end{align*}, except to say that \begin{align*}V\end{align*} must be is at least some minimum value if the intended target is to be reached.

Since \begin{align*}45^{\circ}\end{align*} gives the greatest range, what kind of range will smaller and larger angles than \begin{align*}45^{\circ}\end{align*} give? Forty-five degrees is an optimum condition and we notice that for \begin{align*}\theta=30^{\circ}\end{align*} and \begin{align*}\theta=60^{\circ}\end{align*} (equally distributed about \begin{align*}45^{\circ}\end{align*} we have \begin{align*}2 \theta = 2(30^{\circ})= 60^{\circ}\end{align*}, and \begin{align*}2 \theta = 2(60^{\circ}) = 120^{\circ}\end{align*} which gives \begin{align*}\sin 60^{\circ} = \sin 120^{\circ}\end{align*}. The implication is that the range is the same for angles \begin{align*} (45^{\circ}+ \theta)\end{align*} and \begin{align*} (45^{\circ}- \theta)\end{align*}. This condition can be readily proven with a bit of trigonometry: Is \begin{align*}\sin (2(45^{\circ}+\theta)) = \sin(2(45^{\circ}-\theta))\end{align*} an identity? The distribution gives \begin{align*}\sin(90^{\circ}+2 \theta) = \sin (90^{\circ}-2 \theta)\end{align*}, which after expansion gives: \begin{align*}\cos (2 \theta) = \cos (2 \theta)\end{align*} and confirms the statement is an identity.

Lastly, we consider expressing \begin{align*}y\end{align*} as a function of \begin{align*}x\end{align*}, rather than \begin{align*}t\end{align*}.

By solving equation 1 for \begin{align*}t\end{align*} and substituting the result into equation 2, we have, after recalling \begin{align*}\tan \theta=\frac{\sin \theta}{\cos \theta}\end{align*},

\begin{align*}y_f=\frac{1}{2}g\frac{(\Delta x)^2}{(v \cos \theta)^2} + \Delta x \tan \theta + y_i\end{align*}

If \begin{align*}X_i = 0\end{align*} then \begin{align*}\Delta x\end{align*} above can be replaced with \begin{align*}X\end{align*}, or \begin{align*}R\end{align*}.