Projectile Motion: Difference between revisions
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<b>Additional examples of projectile motion can be found on the [[Kinematics]] page.</b> | <b>Additional examples of projectile motion can be found on the [[Kinematics]] page.</b> | ||
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===Simple=== | |||
A ball of mass <math> m = 1000 \; g </math> rolls across the floor with a velocity of <math> \vec{v} = (3,4,0) \;m/s </math> . After how much time does the ball stop? Where does it stop if it starts at the origin? Assume the coefficient of friction is 0.3. | |||
<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;"> | |||
<div style="font-weight:bold;line-height:1.6;">Solution</div> | |||
<div class="mw-collapsible-content"> | |||
We need to find when the velocity will be equal to zero, given a constant frictional force | |||
'''Declare known variables and coordinate system:''' | |||
First, we convert mass into kilograms | |||
<math> \mathbf{m} = \frac{1000} {1000} = 1 kg </math> | |||
Next, we restate the velocity and coefficient of friction | |||
<math> \vec{\mathbf{v}} = (3,4,0) \; \frac{m} {s} </math> | |||
<math> \mu = .3 </math> | |||
Next, from the context of the problem, we take x and y to be perpendicular to the force of gravity (that is, the floor is flat), so gravity will be directed in the negative z direction. | |||
'''Find the initial momentum:''' | |||
<math> \vec{p}_{i} = \mathbf{m} \vec{\mathbf{v}} = (1 \; kg)((3,4,0) \frac{m}{s}) = (3,4,0) \; \frac{kg \; m}{s}</math> | |||
'''Find time passed:''' | |||
We have the momentum principle: | |||
<math> \Delta{\vec{p}} = \vec{\mathbf{F}}_{net} \Delta t </math> | |||
Next, we compute the force due to friction (see [[Friction]]): | |||
<math> \vec{\mathbf{F}}_{frict} = \vec{\mathbf{F}}_{N}*\mu </math> | |||
Expanding in the coordinate system we defined earlier: | |||
<math> \vec{\mathbf{F}}_{normal} = (0,0,\mathbf{m}|\mathbf{g}|) = (0,0,9.8) N </math> | |||
Now we compute friction. Friction will always be directed in the opposite direction of the velocity, so just as velocity has vector for <math> \vec{\mathbf{v}} = (3,4,0) \; m/s </math>, friction will have vector form <math> \vec{\mathbf{F}}_{frict} = |\vec{\mathbf{F}}_{frict}|(-3/5,-4/5,0) </math>, where <math> |\vec{\mathbf{F}}_{frict}| </math> is the magnitude of the frictional force, and the vector has been normalized, so that the its magnitude is equal to one (you should recognize 3,4, and 5 as a pythagorean triple). Therefore, we calculate | |||
<math> |\vec{\mathbf{F}}_{frict}| = \mu |\vec{\mathbf{F}}_{normal}| = 2.94 \;N </math> | |||
<math> \vec{\mathbf{F}}_{frict} = (2.94 \; N)(-3/5,-4/5,0) = (-1.764,-2.352,0) \; N </math> | |||
Now, we compute the time it takes to stop based on the change in momentum | |||
<math> \Delta \vec{p} = \vec{p}_{i} = (\Delta t)\vec{\mathbf{f}}_{frict} </math> | |||
Since the final momentum will be 0, the change in momentum is simply <math> \Delta \vec{\mathbf{p}} = - \vec{\mathbf{p}}_i </math>, so we have | |||
<math> (-3,-4,0) = (-1.764,-2.352,0) N * \Delta {t} </math> | |||
<math> \Delta {t} = \frac {(3,-4.,0) \; kg \; m/s} {(-1.764,-2.352,0) \; N} </math> | |||
<math> \Delta {t} = 1.7 s </math> | |||
'''Find displacement''' | |||
We have two options. We may use average velocity: | |||
<math> \Delta {d} = \vec{v}_{avg} * \Delta {t} = \frac {(3,4,0)} {2} \frac{m} {s} * 1.7 s = (2.55,3.4,0) m </math> | |||
and then find the final position: | |||
<math> \mathbf{r}_{f} = \mathbf{r}_{i} + \Delta {d} = (0,0,0) m + (2.55,3.4,0) m = (2.55, 3.4,0) m</math> | |||
Alternatively, we may use the kinematics equation | |||
<math> \vec{\mathbf{r}}_f = \frac{\vec{\mathbf{a}} t^2}{2} + \vec{\mathbf{v}}_i t + \vec{\mathbf{r}}_i </math> | |||
where acceleration <math> \vec{a} = \vec{\mathbf{F}}/\mathbf{m} = (-1.764, -2.352, 0) m/s^2 </math>. Thus this yields | |||
<math> \vec{\mathbf{r}}_f = \frac{(-1.764, -2.352, 0)(1.7 \; s)^2}{2} + (3,4,0)(1.7 \; s) + (0,0,0) = (2.55, 3.4, 0) \; m </math> | |||
</div></div> | |||
===Middling=== | |||
<div class="toccolours mw-collapsible mw-collapsed" style="width:900px; overflow:auto;"> | |||
<div style="font-weight:bold;line-height:1.6;">Solution</div> | |||
<div class="mw-collapsible-content"> | |||
You kick a <math>5 \;kg </math> ball off a <math>100\;m</math> cliff at a velocity of <math> (10,15,10)\; m/s </math>. Assume gravity acts in the negative z direction, and that there is no air resistance. How long does it take for the ball to reach the ground? How far away does it land? | |||
'''Declare Known Variables''' | |||
First we have mass <math> \mathbf{m} = 5\; kg </math>. Next, we know that <math> \mathbf{h} = 100\; m </math>, which may be written as <math> \vec{\mathbf{r}}_i = (0,0,100)\; m </math>. fly, we have <math> \vec{\mathbf{v}}_{i} = (10,15,10) \;\frac{m} {s} </math>. | |||
'''Find Initial Momentum''' | |||
Using the definition of [[Linear Momentum]], we find that | |||
<math> \vec{\mathbf{p}}_{i} = \mathbf{m} \vec{\mathbf{v}}_{i} </math> | |||
<math> \vec{\mathbf{p}}_{i} = (5\;kg) (10,15,10) \;\frac{m} {s} = (50,75,50) \; \frac{kg \; m} {s}</math> | |||
'''Find Net Force''' | |||
Because gravity is directed in the z direction, the force will have only a z component, which we may calculate: | |||
<math> \vec{\mathbf{F}}_{net} = \vec{\mathbf{F}}_{g} = (0,\mathbf{m}\mathbf{g},0) = (0, (5\;kg)(9.8 \;\frac{m} {s^2}),0) = (0,49,0)\; N</math> | |||
'''Find time passed''' | |||
We know that the final z coordinate of the ball will be equal to zero, so we compute the duration of the flight by calculating when this will occur. To do this we use kinematics: | |||
<math> z_f = \frac{1}{2} a_z t^2 + v_z t + z_i </math> | |||
<math> 0 = \frac{1}{2} (-9.8 \; \frac{m}{s^2})t^2 + (10 \; \frac{m}{s})t + 100 \; m </math> | |||
This is a standard quadratic equation, so we can use the quadratic formula to find that: | |||
<math> t = \frac{-b \pm \sqrt{b^2-4ac}}{2a} = \frac{ -10 \; \frac{m}{s} \pm \sqrt{100 \; \frac{m^2}{s^2} + 1960 \frac{m^2}{s^2}}}{-9.8 \; \frac{m}{s^2}} </math> | |||
Notice the negative denominator, so it is the minus option which will be positive. Since negative times are unphysical, we discard the plus case and have that <math> t \approx 5.65 s </math>. | |||
'''Find Displacement in x''' | |||
Since no force acts in this direction, it is very straightforward to compute that | |||
<math> x_f = x_i + v_x t</math> | |||
<math> x_f = 0 + (10 \; \frac{m}{s})(5.65 \; s) = 56.5 \; m </math> | |||
'''Find Displacement in y''' | |||
As with above, we may compute | |||
<math> y_f = y_i + v_y t</math> | |||
<math> y_f = 0 + (15 \; \frac{m}{s})(5.65 \; s) = 84.75 \; m</math> | |||
Combining all this, the ball's final position will be <math> (56.5,84,75,0) \; m </math> | |||
</div></div> | |||
===Difficult=== | |||
You kick a ball through 5 meter high posts from 60 meters away. It takes 2.5s to travel through the posts. Answer the following questions: | |||
a)What is the initial velocity? | |||
b)What angle did you shoot it from? | |||
c)What is the velocity of the ball as it crosses the posts? | |||
d)What was the balls maximum height? | |||
e)What was the force on the ball if the ball is .76 kg and the impact lasts for 34 ms? | |||
====Part A==== | |||
'''Find Initial Velocity in X direction''' | |||
<math> \frac{\Delta {p}_x} {\Delta {t}} = \mathbf{F}_{net} = 0 </math> | |||
<math> \frac{m *\Delta {v}_x} {\Delta {t}} = 0 </math> | |||
Or the x component of the velocity is constant. | |||
<math> \overrightarrow{v}_x = \frac{\Delta {x}} {\Delta {t}} = \frac{60} {2.5} = 24 \frac{m}{s} </math> | |||
'''Find Initial Velocity in Y direction''' | |||
<math> \frac{\Delta {p}_y} {\Delta {t}} = \mathbf{F}_{net} = - \mathbf{m}\mathbf{g} = \frac{m *\Delta {v}_y} {\Delta {t}} </math> | |||
<math> \frac{\Delta {v}_y} {\Delta {t}} = -\mathbf{g}</math> | |||
<math> \Delta {v}_y = - \mathbf{g} * \Delta {t} = v_{final y} - v_{initial y}</math> | |||
<math> v_{final y} = - \mathbf{g} * \Delta {t} + v_{initial y}</math> | |||
<math> v_{avg y} = \frac{v_{final y} + v_{initial y}} {2} </math> | |||
<math> v_{final y} = 2v_{avg y} - v_{initial y} </math> | |||
<math>- \mathbf{g} * \Delta {t} + v_{initial y} = 2v_{avg y} - v_{initial y} </math> | |||
<math> v_{avg y} = v_{initial y} - \frac{1} {2} \mathbf{g} \Delta {t} </math> | |||
<math> \frac{\Delta {y}} {\Delta {t}} = v_{avg y} </math> | |||
<math> \Delta {y} = v_{initial y} \Delta {t} - \frac{1} {2} \mathbf{g} \Delta {t}^2</math> | |||
<math> v_{initial y} = \frac{\Delta {y}} {\Delta {t}} + \frac{1} {2} \mathbf{g} \Delta {t} = \frac{5} {2.5} + \frac{1} {2} * 9.8 * 2.5 = 14.25 \frac {m} {s} </math> | |||
'''Find Initial Velocity in Z direction''' | |||
No change in Z direction, therefore: | |||
<math> v_{initial z} = 0 \frac{m} {s} </math> | |||
'''Find Initial Velocity''' | |||
<math> \overrightarrow{v}_{initial} = (24,14.25,0) \frac {m} {s}</math> | |||
====Part B==== | |||
'''Find the Angle''' | |||
<math> \tan {\theta} = \frac{\mathbf{v_{initial y}}} {\mathbf{v_{initial x}}} = \frac{14.25} {24} = 0.59375</math> | |||
<math> \theta = \tan^{-1} 0.59375 = 0.53358 radians</math> or 30.6997 degrees | |||
====Part C==== | |||
'''Find Final Velocities''' | |||
Final velocity is equal to the velocity as it goes through the poles. | |||
No change in x velocity: | |||
<math> v_{final x} = 24 \frac{m} {s} </math> | |||
From part A: | |||
<math> v_{final y} = - \mathbf{g} * \Delta {t} + v_{initial y}</math> | |||
<math> v_{final y} = - 9.8 * 2.5 + 14.25</math> | |||
<math> v_{final y} = - 10.25 \frac{m} {s}</math> | |||
<math>\overrightarrow{v}_{final} = (24, -10.25, 0) \frac{m} {s}</math> | |||
====Part D==== | |||
'''Find Max Time''' | |||
The velocity at the max point is 0. Use equation from part A: | |||
<math> v_{max y} = v_{initial y} - g \Delta{t_{max}} = 0</math> | |||
<math>v_{initial y} = g \Delta{t_{max}}</math> | |||
<math>\Delta{t_{max}} = \frac{v_{initial y}} {g}</math> | |||
'''Find Max Height''' | |||
<math> \frac{\Delta {y}} {\Delta{t}} = v_{initial y} - \frac{1}{2} g \Delta{t_{max}}</math> | |||
<math> y_{max} - y_{initial} = v_{initial y} \Delta{t_{max}} - \frac{1}{2} g \Delta{t_{max}}^2</math> | |||
<math> y_{max} = \frac{v_{initial y}^2} {g} - \frac{1}{2} {(\frac{v_{initial y}^2} {g})}</math> | |||
<math> y_{max} = \frac{1}{2} {(\frac{v_{initial y}^2} {g})} = \frac{1}{2} {(\frac{14.25^2} {9.8})} = 10.3603 m</math> | |||
====Part E==== | |||
'''State Known Variables''' | |||
<math> \mathbf{m} = 0.76 kg </math> | |||
<math> \Delta{t} = .034 s </math> | |||
'''Find Change in Momentum''' | |||
<math> \overrightarrow{\mathbf{p}}_{initial} = 0 </math> | |||
<math> \overrightarrow{\mathbf{p}}_{final} = m * \overrightarrow{v}_{initial} </math> | |||
Use V Initial from Part A: | |||
<math> \Delta{\overrightarrow{\mathbf{p}}} = \overrightarrow{\mathbf{p}}_{final} - \overrightarrow{\mathbf{p}}_{initial} = m * \overrightarrow{v}_{initial} - 0 = 0.76 kg * (24,14.25,0) \frac{m} {s} = (18.24,10.83,0) kg \frac{m} {s}</math> | |||
<math> |\Delta{\mathbf{p}}| = \sqrt{18.24^2 + 10.83^2} = 21.213 kg \frac{m} {s}</math> | |||
'''Find Net Force''' | |||
<math> |\Delta{\mathbf{p}}| = |\mathbf{F}_{net}| * \Delta{t} </math> | |||
<math> |\mathbf{F}_{net}| = \frac{|\Delta{\mathbf{p}}|} {\Delta{t}} = \frac{21.213} {.034} = 623.908 N</math> | |||
--> | |||
==Connectedness== | ==Connectedness== | ||
Revision as of 10:23, 2 July 2019
This page will attempt to analyze projectile motion, a branch of classical mechanics in which the motion of an object (the projectile) is analyzed under the influence of the constant acceleration of gravity, after it has been propelled with some initial velocity.
The Main Idea
Projectile motion is a branch of classical mechanics which analyzes the motion of objects (projectiles) under the influence of the constant acceleration of gravity near the surface of the earth. Often, the projectiles are propelled prior to being analyzed under the influence of gravity, giving them initial velocities that affect their trajectories. Two classic examples of projectile motion are the firing of a cannon and throwing a ball off a tower, but there are many other examples of projectile motion. To be considered a projectile, the only force that can be acting upon the object is gravity. Note that for real projectiles, other forces such as air resistance are usually present, but these are usually negligible in magnitude and make the mathematical model significantly more complicated, so these are ignored. Regardless of the direction of motion of the projectile, the free-body diagram of a projectile is always the same, and constant throughout its trajectory: a particle on which only gravity acts in a downward direction.
A Mathematical Model
Traditionally, the Frame of Reference chosen for projectile motion problems places the origin on the ground below the point of launch and defines t=0 to be the time of launch. The +y direction is usually defined as vertically upwards, so the gravitational force acts in the -y direction. The +x direction is usually defined as the direction of the horizontal component of the projectile's velocity. No forces act in the x direction, so the x component of the projectile's velocity is constant throughout the trajectory. The y component of the velocity is constantly changing over time due to the gravitational force. Gravity near the surface of the earth causes a constant downward acceleration of 9.8m/s^2. This causes a parabolic trajectory. The constant velocity in the x direction and the constant acceleration in the y direction lend themselves to the Kinematic Equations. Kinematics are used to analytically solve problems related to projectile motion.
The following equations derived from kinematics describe the x and y components of a projectile's position as functions of time.
[math]\displaystyle{ y(t) = - \frac{1}{2} g * t^2 + v_{y, 0} * t + y_i }[/math]
[math]\displaystyle{ x(t) = v_{x, 0} * t }[/math]
It is evident from the above equations that the two components of a projectile's motion are independent; the particle's horizontal position has no effect on its vertical motion and vice versa.
[math]\displaystyle{ v_{y, 0} }[/math] and [math]\displaystyle{ v_{y, 0} }[/math] may each be positive, negative, or 0, depending on how the projectile was launched, resulting in a variety of trajectory shapes. Below are some possible trajectories of projectiles:
Note that all of these shapes are parabolic.
Sometimes, an initial speed and an initial angle (usually with respect to the horizontal) is given. In this case, [math]\displaystyle{ v_{y, 0} = v_0 \sin \Theta_0 }[/math] and [math]\displaystyle{ v_{x, 0} = v_0 \cos \Theta_0 }[/math].
Sometimes, rather than giving an initial velocity, a projectile motion problem will give the mass of the projectile as well as the magnitude, direction, and duration of the propelling force launching it. From this information, the relationship between Impulse and Momentum can be used to find the initial velocity of the projectile. In these situations, the propelling force is usually considered the only force acting on the object during the pre-trajectory impulse, because even if other forces are present, they are usually negligible compared to the propelling force.
At the highest point in a projectile's trajectory, the projectile's velocity is entirely horizontal; the vertical component is 0. This is because at that point in time, the vertical velocity has stopped being positive and is no longer carrying the projectile higher, but has not yet become negative and has therefore not begun lowering the projectile once more.
A Computational Model
One method used to visualize or predict a projectiles trajectory is to apply our mathematical model using computational programming. Computer models usually use Iterative Prediction rather than Kinematics to determine the path of the projectile. Here is an example of a program in vPython that simulates a projectile:
Click "view this program" on the top left corner to view the source code.
Examples
1. (Simple)
2. (Middling)
In the example above, the time of flight of each cannonball is determined by finding the time until its highest point and doubling that. This is possible because for this problem the ground is flat; if the cannonball ends up higher or lower than it starts, the time of flight will have to be found using its y-coordinate as a function of time. Once the time of flight is found, it is multiplied by the x component of its velocity to find its horizontal displacement.
As a challenge, consider trying to prove that 45[math]\displaystyle{ ^\circ }[/math] gives a greater range than any other angle, rather than comparing it to only 30[math]\displaystyle{ ^\circ }[/math] and 60[math]\displaystyle{ ^\circ }[/math]. Hint: find the range of the projectile as a function of the launch angle, then use optimization (show that the derivative of the function is 0 at [math]\displaystyle{ \Theta = }[/math] 45[math]\displaystyle{ ^\circ }[/math] and that therefore a maximum occurs there).
Additional examples of projectile motion can be found on the Kinematics page.
Connectedness
Projectile motion has many applications in day-to-day life; every dropped, thrown, and launched object massive enough to disregard air friction is an example of projectile motion.
Application: rifle scope industry
A bullet fired from a horizontal hunting rifle obeys the model of projectile motion, so by the time it reaches a given horizontal distance from the hunter, it will have fallen vertically by a predictable amount. The scope of a hunting rifle can be calibrated to specific distances. This angles the scope slightly below the horizontal to take into account the bullet drop; when the bullet reaches the horizontal distance of interest, it will have fallen exactly into the crosshair of the scope. The design and manufacture of the scopes and their calibration mechanisms require a quantitative understanding of projectile motion.
History
Several important figures have contributed to the modern understanding of projectile motion.
Aristotle
The first findings of Projectile Motion were contributed by Aristotle. He believed that a projectile modeled by a canon ball travels in a straight line until it lost its 'impetus'. When it lost its 'impetus' which ran out after time, due to a restive force, the projectile would immediately drop and fall to the ground. For objects projected over short distances, at first observation this can visually appear to be true.
Galileo
Galileo's findings were motivated by the Renaissance. Through as series of experiments, he discovered that projectile motion is a result of free fall motion along the y-axis and uniform (constant velocity) motion along x-axis. This model became well-known and led to new insights on projectile motion, such as its parabolic shape. It is still accepted today.
One experiment Galileo performed in developing this model consisted of dropping two objects from equal heights: one from rest, and one with an initial horizontal velocity. Both struck the ground at the same time because both had the same initial vertical velocity. The horizontal component of velocity does not affect its vertical motion because the components of motion are independent.
Isaac Newton
Isaac Newton contributed to the idea of projectile motion with his idea of a cannon on a mountaintop, now known as Newton's cannonball. Newton hypothesized that a cannon on top of a high mountain could fire a ball with varying initial speeds for varying trajectories. A cannonball simply dropped off the mountain would merely return to earth, but if that cannonball were fired from a cannon then it would return to Earth in an arc, or parabolic shape. This would be considered projectile motion. Neither of these scenarios were new or particularly innovative, but Newton also imagined a third scenario in which the cannonball were launched at great speed. In this scenario, he imagined that the ball would travel so far before curving downward that the earth would curve underneath it due to its spherical shape, resulting in a circular orbit. In doing so, Isaac Newton became one of the first to imagine universal gravity and to come up with a cause for orbital motion. It is worth noting that Newton's contribution does not strictly fall under the category of projectile motion as it is defined on this page because in the case of a cannonball orbiting the earth, gravity is not constant and downward, but changes its direction depending on the cannonball's position to point towards the center of the earth.
See also
External links
Interactive Projectile Simulator
References
https://www.phy.duke.edu/~rgb/Class/intro_physics_1/intro_physics_1.pdf
http://www.physicsclassroom.com/class/vectors/Lesson-2/What-is-a-Projectile
https://vidphysics.blogspot.ca/p/blog-page.html
http://blogs.umass.edu/ndarnton/2009/02/28/projectile-motion-in-biology/