Projectiles

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© 1998-2008 by Glenn Elert -- A Work in Progress
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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 …

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 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 ± 0.5 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 further than the radius of metropolitan Los Angeles, 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 − ½ gt2   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 =  (v cos θ) t     y =  y0  +  v0yt  −  ½ gt2
                   
xfinal =  (v cos θ) tfinal   0 =  0  +  (v sin θ)tfina  −  ½ gt2
                   
xfinal =  (v cos θ) 2(v sin θ)   tfinal =      2(v sin θ)    
g       g    
xfinal =  v2 sin 2θ xmax =  v2            
g g            

Summary

Problems

practice

  1. In the 1994 action-adventure film Speed, an extortionist equipped a Los Angeles bus with a bomb that would explode if the speed of the bus fell below 50 mph (22 m/s). The police discovered the bomb and routed the bus on to a segment of freeway that was still under construction -- their intention being to keep it out of the notoriously heavy Southern California traffic. In a twist of the plot, however, they overlooked a 50 foot (15 m) gap in construction. Since the bomb would explode killing everyone on board if they slowed down, they decide to jump the gap. The missing segment lies on an effectively level stretch of elevated freeway, 50 feet (15 m) above the ground, yet somehow they manage to jump the gap at 67 mph (30 m/s) and land safely on the other side.
             
       
             
    1. If the bus in Speed obeyed the laws of physics how would this scene end?
    2. Just out of curiosity, what would happen if the bus in the movie Speed was odriven off of a 15 m high, horizontal ramp?
      1. Where would the bus land?
      2. What velocity would it have when it struck the ground?
    Solutions …
    1. Although the bus is accelerating downward due to gravity its horizontal velocity remains constant. (Gravity never acts horizontally.) It will take the bus …
               
      Δt =  Δx  =  15 m  = 0.50 s
      v 30 m/s
               
      to travel the width of the gap, during which time it would fall …
               
      y =  1  at2 =  1  ( 9.8 m/s2)(0.50 s)2 = 1.3 m
      2 2
               
      This means the bus will strike the flat concrete face of the freeway stub somewhere above its front bumper. So much for Speed II.
    2. Let's let the bus fly free!
      1. The first part of this question is basically asking how far forward a bus moving at 30 m/s would travel in the time it took for it to fall 15 m downward. In this problem there are two independent equations of motion -- one with constant velocity (the horizontal motion) and one with constant acceleration (the vertical motion). Since the ramp is horizontal, there is no initial velocity in the vertical direction. Thus, the time it takes a horizontally launched projectile to reach the ground is the same as the time it takes an object released from rest to fall the same height.
                 
        y0 =  0 m   y =  y0 + v0yΔt + ½ayΔt2
        y =  15 m   y =  ½ayΔt2
        v0y =  0 m/s   Δt =  √(2y/ay) = √[2(15 m)(9.8 m/s2)]
        ay =  9.8 m/s2   Δt =  1.75 s
                 
        Now use the time of flight to determine the horizontal displacement of the bus as it flies through the air.
                 
        vx =  30 m/s   x =  vxΔt
        Δt =  1.75 s   x =  (30 m/s)(1.75 s)
              x =  53 m
                 
        Look for the bus 53 m in front of a point directly below the edge of the ramp.
      2. The final velocity of the bus will be the vector sum of its horizontal and vertical velocities on impact. Since there is no horizontal acceleration, the velocity in this direction remains constant. If the bus leaves the ramp at 30 m/s horizontally it will strike the ground at 30 m/s horizontally. There is an acceleration vertically, however. Initially, the bus has no vertical velocity, but this will obviously change. Gravity pulls everything relentlessly toward the earth. Once there's no more road to hold up the bus, there's nothing to keep it from accelerating down.
                 
        y0 =  0 m   vy2 =  voy2 + 2ayy
        y =  15 m   vy =  √(2ayy)
        v0y =  0 m/s   vy =  √[2(9.8 m/s2)(15 m)]
        ay =  9.8 m/s2   vy =  17.1 m/s
                 
        Now that we have both components, find the magnitude and direction of the resultant velocity by the usual means: Pythagorean theorem and the definition of the tangent function.
                       
        v2 =  vx2 + vy2   tan θ =  vy  =  17.1 m/s  
        v2 =  (30 m/s)2 + (17.1 m/s)2   vx 30 m/s  
        v =  35 m/s   θ =  30°(angle of depression)
                       
  2. Some sort of projectile launched at an angle would be good here -- perhaps the demo done in class. Answer it.
  3. Shoot the monkey
     
    [magnify]
     
    • Describe the solution.
  4. Write something completely different.
    • Answer it.

conceptual

  1. When would a bird in the air not be considered a projectile? Under what circumstances could that same bird now become a projectile?

numerical

  1. While driving on the interstate one day at 27.8 m/s (60.0 mph) I accidentally dropped the Encyclopedia of Physics out the window, 1.15 m above the ground. Determine the following …
    1. the horizontal and vertical components of the book's velocity the instant I released it
    2. the time the book was in the air
    3. the horizontal distance the book traveled before hitting the ground
    4. the horizontal and vertical components of the book's velocity the instant it hit the ground
  2. A ball rolls off the top step of a long staircase as shown in the diagram below. The rise and run of each step are 16 and 32 cm respectively.
     
    [magnify]
     
    Determine …
    1. the step on which the ball lands if it rolled off the top step at 5.00 m/s
    2. the ball's speed when it left the top step (stated as a range) if it landed on the seventh step
  3. At the 1998 Punkin Chunkin World Championship, a pneumatically-driven device called the "Aludium Q36 Pumpkin Modulator" was able to project an 3.6-4.5 kg (8-10 pounds) pumpkin intact for a total distance of 1227.23 m (4026.32 feet). What was the muzzle velocity (magnitude and direction) of the record-setting pumpkin?
  4. Incomplete. Jackie Chan in Rumble in the Bronx. As Mr. Chan's stunt adviser you must tell him how fast to run. What is your answer?
  5. Incomplete. For reasons beyond the scope of this book, the optimal launch angle for a long jumper is 35.26° Bob Beamon’s spectacular long jump of 8.90m would last as a world record for 22 years. Mexico City 1968 Games of the 19th Olympiad. 350.5 inches or 8.90 m. What was Bob Beamon's take off speed?

calculus

  1. Incomplete. Prove the following …
    1. max distance at 45°
    2. max path length at 56.46°
    3. max area under curve at 60°
    4. max flight time at 90°

investigative

  1. How high should a domed stadium be built so that no batted ball would ever strike the ceiling? Solve usinng the distance of the farthest batted ball.

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