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From: Mary John <elecncomp@gmail.com>
Date: Tue, Oct 25, 2011 at 4:09 PM
Subject: Fwd: TEC 105
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From: Mary John <elecncomp@gmail.com>
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Subject: Fwd: TEC 105
To: ivy.german@yahoo.com
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Subject: Fwd: TEC 105
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Subject: Fwd: TEC 105
From: Mary John <elecncomp@gmail.com>
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Subject: Fwd: TEC 105
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Subject: Fwd: TEC 105
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From: obadiah maube
Date: Saturday, October 22, 2011
Subject: TEC 105
To: "pius.nondi@gmail.com" <pius.nondi@gmail.com>, "jackybosi@gmail.com" <
jackybosi@gmail.com>, "iamme_mochs@yahoo.com" <iamme_mochs@yahoo.com>, "
kiruisimoo@yahoo.com" <kiruisimoo@yahoo.com>
Find attached notes for TEC 105. ensure your colleagues have the notes
before the next lesson.
Regards Maube
Find attached notes for TEC 105. More to follow. Ensure that you deliver
them to your colleagues.
regards
Maube
From: obadiah maube
Date: Saturday, October 22, 2011
Subject: TEC 105
To: "pius.nondi@gmail.com" <pius.nondi@gmail.com>, "jackybosi@gmail.com" <
jackybosi@gmail.com>, "iamme_mochs@yahoo.com" <iamme_mochs@yahoo.com>, "
kiruisimoo@yahoo.com" <kiruisimoo@yahoo.com>
Find attached notes for TEC 105. ensure your colleagues have the notes
before the next lesson.
Regards Maube
Find attached notes for TEC 105. More to follow. Ensure that you deliver
them to your colleagues.
regards
Maube
RANGE OF PROJECTILE
In physics, assuming a flat Earth with a uniform gravity field, a projectile launched with specific initial conditions will have a predictable range. As in Trajectory of a projectile, we will use:
The following applies for ranges which are small compared to the size of the Earth. For longer ranges see sub-orbital spaceflight.
g: the gravitational acceleration—usually taken to be 9.81 m/s2 (32 f/s2) near the Earth's surface
θ: the angle at which the projectile is launched
v: the velocity at which the projectile is launched
y0: the initial height of the projectile
d: the total horizontal distance travelled by the projectile
When neglecting air resistance, the range of a projectile will be
d = \frac{v \cos \theta}{g} \left( v \sin \theta + \sqrt{(v \sin \theta)^2 + 2gy_0} \right)
If (y0) is taken to be zero, meaning the object is being launched on flat ground, the range of the projectile will then simplify to
d = \frac{v^2}{g} \sin 2 \theta
Contents
[hide]
1 Ideal projectile motion
1.1 Derivations
1.1.1 Flat Ground
1.1.2 Uneven Ground
1.1.3 Maximum Range
1.1.4 Angle of impact
2 Actual projectile motion
2.1 Projectile characteristics
2.2 Firearm barrels
3 See also
[edit] Ideal projectile motion
Ideal projectile motion assumes that there is no air resistance. This assumption simplifies the math greatly, and is a close approximation of actual projectile motion in cases where the distances travelled are small. Ideal projectile motion is also a good introduction to the topic before adding the complications of air resistance.
[edit] Derivations
[edit] Flat Ground
Range of a projectile (in vacuum).
First we examine the case where (y0) is zero. The horizontal position of the projectile is
x(t) = \frac{}{} v\cos \left(\theta\right) t
In the vertical direction
y(t) = \frac{} {} v\sin \left(\theta\right) t - \frac{1} {2} g t^2
We are interested in the time when the projectile returns to the same height it originated at, thus
0 = \frac{} {} v\sin \left(\theta\right) t - \frac{1} {2} g t^2
By factoring:
\frac{} {}t = 0
or
t = \frac{2 v \sin \theta} {g}
The first solution corresponds to when the projectile is first launched. The second solution is the useful one for determining the range of the projectile. Plugging this value for (t) into the horizontal equation yields
x = \frac {2 v^2 \cos \left(\theta\right) \sin \left(\theta\right)} {g}
Applying the trigonometric identity
sin(x + y) = sin(x)cos(y) + sin(y)cos(x)
If x and y are same,
sin(2x) = 2sin(x)cos(x)
allows us to simplify the solution to
d = \frac {v^2} {g} \sin \left(2 \theta\right)
Note that when (θ) is 45°, the solution becomes
d = \frac {v^2} {g}
[edit] Uneven Ground
Now we will allow (y0) to be nonzero. Our equations of motion are now
x(t) = \frac{}{} v\cos \left(\theta\right) t
and
y(t) = y_0 + \frac{} {} v\sin \left(\theta\right) t - \frac{1} {2} g t^2
Once again we solve for (t) in the case where the (y) position of the projectile is at zero (since this is how we defined our starting height to begin with)
0 = y_0 + \frac{} {} v\sin \left(\theta\right) t - \frac{1} {2} g t^2
Again by applying the quadratic formula we find two solutions for the time. After several steps of algebraic manipulation
t = \frac {v \sin \theta} {g} \pm \frac {\sqrt{\left(v \sin \theta\right)^2 + 2 g y_0}} {g}
The square root must be a positive number, and since the velocity and the cosine of the launch angle can also be assumed to be positive, the solution with the greater time will occur when the positive of the plus or minus sign is used. Thus, the solution is
t = \frac {v \sin \theta} {g} + \frac {\sqrt{\left(v \sin \theta\right)^2 + 2 g y_0}} {g}
Solving for the range once again
d = \frac {v \cos \theta} {g} \left [ v \sin \theta + \sqrt{\left(v \sin \theta \right)^2 + 2 g y_0} \right]
[edit] Maximum Range
Trajectories of projectiles launched at different elevation angles but the same speed of 10 m/s in a vacuum and uniform downward gravity field of 10 m/s2. Points are at 0.05 s intervals and length of their tails is linearly proportional to their speed. t = time from launch, T = time of flight, R = range and H = highest point of trajectory (indicated with arrows).
For cases where the projectile lands at the same height from which it is launched, the maximum range is obtained by using a launch angle of 45 degrees. A projectile that is launched with an elevation of 0 degrees will strike the ground immediately (range = 0), though it may then bounce or roll. A projectile that is fired with an elevation of 90 degrees (i.e. straight up) will travel straight up, then straight down, and strike the ground at the point from which it is launched, again yielding a range of 0.
The elevation angle which will provide the maximum range when launching the projectile from a non-zero initial height can be computed by finding the derivative of the range with respect to the elevation angle and setting the derivative to zero to find the extremum:
\frac { dR } { d\theta} = \frac {v^2} {g} [ \cos \theta ( \cos \theta + \frac {\sin \theta \cdot \cos \theta} {\sqrt {(\sin \theta)^2 + C} } ) - \sin \theta ( \sin \theta + \sqrt { ( \sin \theta )^2 + C })]
where C = \frac {2 g y_0} { v^2} and \,\!R = horizontal range.
Setting the derivative to zero provides the equation:
(\cos \theta)^2 + \frac {\sin \theta \cdot (\cos \theta )^2} {\sqrt {(\sin \theta )^2 + C} } - ( \sin \theta )^2 - \sin \theta \sqrt {(\sin \theta )^2 + C} = 0
Substituting u = (cos θ)2 and 1 − u = (sin θ)2 produces:
u + \frac {u \sqrt {1-u}} {\sqrt {1-u+C}} - (1-u) - (\sqrt {1-u}) \sqrt {1-u+C} = 0
Which reduces to the extremely surprisingly simple expression:
u = \frac {C + 1} {C + 2}
Replacing our substitutions yields the angle that produces the maximum range for uneven ground, ignoring air resistance:
\theta = \arccos \sqrt { \frac {2 g y_0 + v^2} {2 g y_0 + 2 v^2} }
Note that for zero initial height, the elevation angle that produces maximum range is 45 degrees, as expected. For positive initial heights, the elevation angle is below 45 degrees, and for negative initial heights (bounded below by y0 > − 0.5v2 / g), the elevation angle is greater than 45 degrees.
Example: For the values g = 9.81m / s2, y0 = 40m , and v = 50m / s, an elevation angle θ = 41.1° produces a maximum range of Rmax = 292.1m.
[edit] Angle of impact
The angle ψ at which the projectile lands is given by:
\tan \psi = \frac {-v_y(t_d)} {v_x(t_d)} = \frac {\sqrt { v^2 (\sin \theta)^2 + 2 g y_0 }} { v \cos \theta}
For maximum range, this results in the following equation:
(\tan \psi)^2 = \frac { 2 g y_0 + v^2 } { v^2 } = C+1
Rewriting the original solution for θ, we get:
(\tan \theta)^2 = \frac { 1 - (\cos \theta)^2 } { (\cos \theta)^2 } = \frac { v^2 } { 2 g y_0 + v^2 } = \frac { 1 } { C + 1 }
Multiplying with the equation for (tan ψ)^2 gives:
(\tan \psi)^2 (\tan \theta)^2 = \frac { 2 g y_0 + v^2 } { v^2 } \frac { v^2 } { 2 g y_0 + v^2 } = 1
Because of the trigonometric identity
\tan (\theta + \psi) = \frac { \tan \theta + \tan \psi } { 1 - \tan \theta \tan \psi } ,
this means that θ + ψ must be 90 degrees.
[edit] Actual projectile motion
In addition to air resistance, which slows a projectile and reduces its range, many other factors also have to be accounted for when actual projectile motion is considered.
[edit] Projectile characteristics
Generally speaking, a projectile with greater volume faces greater air resistance, reducing the range of the projectile. This can be modified by the projectile shape: a tall and wide, but short projectile will face greater air resistance than a low and narrow, but long, projectile of the same volume. The surface of the projectile also must be considered: a smooth projectile will face less air resistance than a rough-surfaced one, and irregularities on the surface of a projectile may change its trajectory if they create more drag on one side of the projectile than on the other. Mass also becomes important, as a more massive projectile will have more kinetic energy, and will thus be less affected by air resistance. The distribution of mass within the projectile can also be important, as an unevenly weighted projectile may spin undesirably, causing irregularities in its trajectory due to the magnus effect.
If a projectile is given rotation along its axes of travel, irregularities in the projectile's shape and weight distribution tend to be canceled out. See rifling for a greater explanation.
[edit] Firearm barrels
For projectiles that are launched by firearms and artillery, the nature of the gun's barrel is also important. Longer barrels allow more of the propellant's energy to be given to the projectile, yielding greater range. Rifling, while it may not increase the average (arithmetic mean) range of many shots from the same gun, will increase the accuracy and precision of the gun.
The following applies for ranges which are small compared to the size of the Earth. For longer ranges see sub-orbital spaceflight.
g: the gravitational acceleration—usually taken to be 9.81 m/s2 (32 f/s2) near the Earth's surface
θ: the angle at which the projectile is launched
v: the velocity at which the projectile is launched
y0: the initial height of the projectile
d: the total horizontal distance travelled by the projectile
When neglecting air resistance, the range of a projectile will be
d = \frac{v \cos \theta}{g} \left( v \sin \theta + \sqrt{(v \sin \theta)^2 + 2gy_0} \right)
If (y0) is taken to be zero, meaning the object is being launched on flat ground, the range of the projectile will then simplify to
d = \frac{v^2}{g} \sin 2 \theta
Contents
[hide]
1 Ideal projectile motion
1.1 Derivations
1.1.1 Flat Ground
1.1.2 Uneven Ground
1.1.3 Maximum Range
1.1.4 Angle of impact
2 Actual projectile motion
2.1 Projectile characteristics
2.2 Firearm barrels
3 See also
[edit] Ideal projectile motion
Ideal projectile motion assumes that there is no air resistance. This assumption simplifies the math greatly, and is a close approximation of actual projectile motion in cases where the distances travelled are small. Ideal projectile motion is also a good introduction to the topic before adding the complications of air resistance.
[edit] Derivations
[edit] Flat Ground
Range of a projectile (in vacuum).
First we examine the case where (y0) is zero. The horizontal position of the projectile is
x(t) = \frac{}{} v\cos \left(\theta\right) t
In the vertical direction
y(t) = \frac{} {} v\sin \left(\theta\right) t - \frac{1} {2} g t^2
We are interested in the time when the projectile returns to the same height it originated at, thus
0 = \frac{} {} v\sin \left(\theta\right) t - \frac{1} {2} g t^2
By factoring:
\frac{} {}t = 0
or
t = \frac{2 v \sin \theta} {g}
The first solution corresponds to when the projectile is first launched. The second solution is the useful one for determining the range of the projectile. Plugging this value for (t) into the horizontal equation yields
x = \frac {2 v^2 \cos \left(\theta\right) \sin \left(\theta\right)} {g}
Applying the trigonometric identity
sin(x + y) = sin(x)cos(y) + sin(y)cos(x)
If x and y are same,
sin(2x) = 2sin(x)cos(x)
allows us to simplify the solution to
d = \frac {v^2} {g} \sin \left(2 \theta\right)
Note that when (θ) is 45°, the solution becomes
d = \frac {v^2} {g}
[edit] Uneven Ground
Now we will allow (y0) to be nonzero. Our equations of motion are now
x(t) = \frac{}{} v\cos \left(\theta\right) t
and
y(t) = y_0 + \frac{} {} v\sin \left(\theta\right) t - \frac{1} {2} g t^2
Once again we solve for (t) in the case where the (y) position of the projectile is at zero (since this is how we defined our starting height to begin with)
0 = y_0 + \frac{} {} v\sin \left(\theta\right) t - \frac{1} {2} g t^2
Again by applying the quadratic formula we find two solutions for the time. After several steps of algebraic manipulation
t = \frac {v \sin \theta} {g} \pm \frac {\sqrt{\left(v \sin \theta\right)^2 + 2 g y_0}} {g}
The square root must be a positive number, and since the velocity and the cosine of the launch angle can also be assumed to be positive, the solution with the greater time will occur when the positive of the plus or minus sign is used. Thus, the solution is
t = \frac {v \sin \theta} {g} + \frac {\sqrt{\left(v \sin \theta\right)^2 + 2 g y_0}} {g}
Solving for the range once again
d = \frac {v \cos \theta} {g} \left [ v \sin \theta + \sqrt{\left(v \sin \theta \right)^2 + 2 g y_0} \right]
[edit] Maximum Range
Trajectories of projectiles launched at different elevation angles but the same speed of 10 m/s in a vacuum and uniform downward gravity field of 10 m/s2. Points are at 0.05 s intervals and length of their tails is linearly proportional to their speed. t = time from launch, T = time of flight, R = range and H = highest point of trajectory (indicated with arrows).
For cases where the projectile lands at the same height from which it is launched, the maximum range is obtained by using a launch angle of 45 degrees. A projectile that is launched with an elevation of 0 degrees will strike the ground immediately (range = 0), though it may then bounce or roll. A projectile that is fired with an elevation of 90 degrees (i.e. straight up) will travel straight up, then straight down, and strike the ground at the point from which it is launched, again yielding a range of 0.
The elevation angle which will provide the maximum range when launching the projectile from a non-zero initial height can be computed by finding the derivative of the range with respect to the elevation angle and setting the derivative to zero to find the extremum:
\frac { dR } { d\theta} = \frac {v^2} {g} [ \cos \theta ( \cos \theta + \frac {\sin \theta \cdot \cos \theta} {\sqrt {(\sin \theta)^2 + C} } ) - \sin \theta ( \sin \theta + \sqrt { ( \sin \theta )^2 + C })]
where C = \frac {2 g y_0} { v^2} and \,\!R = horizontal range.
Setting the derivative to zero provides the equation:
(\cos \theta)^2 + \frac {\sin \theta \cdot (\cos \theta )^2} {\sqrt {(\sin \theta )^2 + C} } - ( \sin \theta )^2 - \sin \theta \sqrt {(\sin \theta )^2 + C} = 0
Substituting u = (cos θ)2 and 1 − u = (sin θ)2 produces:
u + \frac {u \sqrt {1-u}} {\sqrt {1-u+C}} - (1-u) - (\sqrt {1-u}) \sqrt {1-u+C} = 0
Which reduces to the extremely surprisingly simple expression:
u = \frac {C + 1} {C + 2}
Replacing our substitutions yields the angle that produces the maximum range for uneven ground, ignoring air resistance:
\theta = \arccos \sqrt { \frac {2 g y_0 + v^2} {2 g y_0 + 2 v^2} }
Note that for zero initial height, the elevation angle that produces maximum range is 45 degrees, as expected. For positive initial heights, the elevation angle is below 45 degrees, and for negative initial heights (bounded below by y0 > − 0.5v2 / g), the elevation angle is greater than 45 degrees.
Example: For the values g = 9.81m / s2, y0 = 40m , and v = 50m / s, an elevation angle θ = 41.1° produces a maximum range of Rmax = 292.1m.
[edit] Angle of impact
The angle ψ at which the projectile lands is given by:
\tan \psi = \frac {-v_y(t_d)} {v_x(t_d)} = \frac {\sqrt { v^2 (\sin \theta)^2 + 2 g y_0 }} { v \cos \theta}
For maximum range, this results in the following equation:
(\tan \psi)^2 = \frac { 2 g y_0 + v^2 } { v^2 } = C+1
Rewriting the original solution for θ, we get:
(\tan \theta)^2 = \frac { 1 - (\cos \theta)^2 } { (\cos \theta)^2 } = \frac { v^2 } { 2 g y_0 + v^2 } = \frac { 1 } { C + 1 }
Multiplying with the equation for (tan ψ)^2 gives:
(\tan \psi)^2 (\tan \theta)^2 = \frac { 2 g y_0 + v^2 } { v^2 } \frac { v^2 } { 2 g y_0 + v^2 } = 1
Because of the trigonometric identity
\tan (\theta + \psi) = \frac { \tan \theta + \tan \psi } { 1 - \tan \theta \tan \psi } ,
this means that θ + ψ must be 90 degrees.
[edit] Actual projectile motion
In addition to air resistance, which slows a projectile and reduces its range, many other factors also have to be accounted for when actual projectile motion is considered.
[edit] Projectile characteristics
Generally speaking, a projectile with greater volume faces greater air resistance, reducing the range of the projectile. This can be modified by the projectile shape: a tall and wide, but short projectile will face greater air resistance than a low and narrow, but long, projectile of the same volume. The surface of the projectile also must be considered: a smooth projectile will face less air resistance than a rough-surfaced one, and irregularities on the surface of a projectile may change its trajectory if they create more drag on one side of the projectile than on the other. Mass also becomes important, as a more massive projectile will have more kinetic energy, and will thus be less affected by air resistance. The distribution of mass within the projectile can also be important, as an unevenly weighted projectile may spin undesirably, causing irregularities in its trajectory due to the magnus effect.
If a projectile is given rotation along its axes of travel, irregularities in the projectile's shape and weight distribution tend to be canceled out. See rifling for a greater explanation.
[edit] Firearm barrels
For projectiles that are launched by firearms and artillery, the nature of the gun's barrel is also important. Longer barrels allow more of the propellant's energy to be given to the projectile, yielding greater range. Rifling, while it may not increase the average (arithmetic mean) range of many shots from the same gun, will increase the accuracy and precision of the gun.
PROJECTILES
Discussion
introduction
A projectile is any object that is cast, fired, flung, heaved, hurled, pitched, tossed, or thrown. (This is an informal definition.) The path of a projectile is called its trajectory. Some examples of projectiles include …
- a baseball that has been pitched, batted, or thrown
- a bullet the instant it exits the barrel of a gun or rifle
- a bus driven off an uncompleted bridge
- a moving airplane in the air with its engines and wings disabled
- a runner in mid stride (since they momentarily lose contact with the ground)
- the space shuttle or any other spacecraft after main engine cut off (MECO)
The force of primary importance acting on a projectile is gravity. This is not to say that other forces do not exist, just that their effect is minimal in comparison. A tossed helium-filled balloon is not normally considered a projectile as the drag and buoyant forces on it are as significant as the weight. Helium-filled balloons can't be thrown long distances and don't normally fall. In contrast, a crashing airplane would be considered a projectile. Even though the drag and buoyant forces acting on it are much greater in absolute terms than they are on the balloon, gravity is what really drives a crashing airplane. The normal amounts of drag and buoyancy just aren't large enough to save the passengers on a doomed flight from an unfortunate end. A projectile is any object with an initial non-zero, horizontal velocity whose acceleration is due to gravity alone.
An essential characteristic of a projectile is that its future has already been preordained. Batters may apply "body English" after hitting a long ball, but they do so strictly for psychological reasons. No amount of leaning to one side will make a foul ball turn fair. Of course, the pilot of a disabled airplane may regain control before crashing and avert disaster, but then the airplane wouldn't be a projectile anymore. An object ceases to be a projectile once any real effect is made to change its trajectory. The trajectory of a projectile is thus entirely determined the moment it satisfies the definition of a projectile.
The only relevant quantities that might vary from projectile to projectile then are initial velocity and initial position
This is where we run into some linguistic complications. Airplanes, guided missiles, and rocket-propelled spacecraft are sometimes also said to follow a trajectory. Since these devices are acted upon by the lift of wings and the thrust of engines in addition to the force of gravity, they are not really projectiles. To get around this dilemma, it is common to use the term ballistic trajectory when dealing with projectiles. The word ballistic has its origins in the Greek word βαλλω (vallo), to throw, and surfaces repeatedly in the technical jargon of weaponry from ancient to modern times. For example …
- The ballista, which looks something like a giant crossbow, was a siege engine used in medieval times to hurl large stones, flaming bundles, infected animal carcasses, and severed human heads into fortifications. Before the invention of gunpowder, ballistas (and catapults and trêbuchets) were the weapons of choice for conquerors.
- An intercontinental ballistic missile is a device for delivering nuclear warheads over long distances. At the start of its journey an ICBM is guided by a rocket engine and stabilizer fins, but soon thereafter it enters the phase of its journey where it is effectively in free fall, traveling fast enough to keep it above the earth's atmosphere for awhile but not fast enough to enter orbit permanently. The adjective "intercontinental" refers to the long range capabilities, while the largely free fall journey it takes makes it "ballistic". ICBMs are the ultimate killing machines, but they have never been used in combat to date.
The wide geographic range as well as the wide historic range of these things we call projectiles raises some problems for the typical student of physics. When a projectile is sent on a very long journey, as is the case with ICBMs, the magnitude and direction of the acceleration due to gravity changes. Gravity isn't constant to begin with, but the effect is not very pronounced over everyday ranges in altitude. From the deepest mines in South Africa to the highest altitudes traversed by commercial airplanes, the magnitude of the acceleration due to gravity is always effectively 9.8 m/s2 ± 0.05 m/s2. Similarly, unless you routinely travel medium to long distances, you aren't likely to experience much of a change in the direction of gravity either. To experience a 1° shift in "down" would require traveling 1/360 of the circumference of the earth — roughly 110 km (70 mi) or the length of a typical morning commute to work in Southern California. Thus for projectiles that won't rise higher than an airplane nor travel farther than the diameter of L.A., gravity is effectively constant. This covers the first five of the examples described at the beginning of this section (baseballs, bullets, buses in action-adventure movies, distressed airplanes, and joggers) but not the sixth (the space shuttle after MECO).
To distinguish such simple projectiles from those where variations in gravity and the curvature of the earth are significant, I propose using the term simple projectile. For the remaining problems, the term general projectile seems appropriate since a general solution in mathematics is one that also includes the special cases, but I'm less adamant about this term.
Consider an effectively spherical earth with a single tall mountain sticking out of it like a giant tumor. Now imagine using this location as a place to launch projectiles horizontally with varying initial velocities. What effect would velocity have on range? Well obviously fast projectiles will travel farther than slow ones. A basic concept associated with speed is that "faster means farther", but the relationship is only approximately linear on a spherical earth. For awhile, doubling speed would mean doubling distance, but eventually the curvature of the earth would start to mess things up. At some speed our hypothetical projectile would make it a quarter of the way around the earth and then half way around and then eventually all the way around. At this point our general projectile ceases to be an object with a launch point and a landing point and it starts being a satellite, permanently circling the earth, perpetually changing direction and thus accelerating under the influence of gravity, but never landing anywhere. Technically, such an object would still be a general projectile, since gravity is the primary source of its acceleration, but somehow this doesn't seem right. Objects traveling through what we call "outer space" hardly seem like projectiles any more. They seem like they reside more in the realm of celestial mechanics than terrestrial mechanics. Such distinctions are arbitrary, however, as there is only one mechanics. The laws of physics are assumed universal until it can be demonstrated otherwise. The unification of physical law is a theme that surfaces from time to time in physics.
A projectile and a satellite are both governed by the same physical principles even though they have different names. A simple projectile is made mathematically simple by an idealization (basically a lie of convenience). By assuming a constant value for the acceleration due to gravity, we make the problem easier to solve and (in many cases) do not really lose all that much in the way of accuracy.
Every projectile problem is essentially two one-dimensional motion problems …
The kinematic equations for a simple projectile are those of an object traveling with constant horizontal velocity and constant vertical acceleration.
| horizontal | vertical | |||
| ax | = 0 | ay | = −g | acceleration |
| vx | = v0x | vy | = v0y − gt | velocity-time |
| x | = x0 + v0xt | y | = y0 + v0yt − ½gt2 | displacement-time |
| vy2 | = v0y2 − 2g(y − y0) | velocity-displacement | ||
finish …
The Trajectory of a Simple Projectile is a Parabola
max range at 45°, equal ranges for launch angles that exceed and fall short of 45° by equal amounts (ex. 40° & 50°, 30° & 60°, 0° & 90°)
| x = | x0 + v0xt + ½axt2 | y | = y0 + v0yt + ½ayt2 | ||||
| x = | 0 + (v cos θ) t + 0 | y | = y0 + (v sin θ)t − ½gt2 | ||||
| 0 | = 0 + (v sin θ)tfinal − ½gt2final | ||||||
| xfinal = | (v cos θ) tfinal |
| |||||
| ⇓ | ⇙ | ||||||
| xfinal = | (v cos θ) | 2(v sin θ) | |||||
| g | |||||||
| xfinal = | v2 sin 2θ | ||||||
| g | |||||||
| xmax = | v2 | ||||||
| g | |||||||