1. Blazing Quadratics Simple Targets Barrier Radar Radar & Barrier
Copyright 2007 by Paul Barber
Simple Targets
Barrier
Radar
Radar & Barrier
Elevated Target
Incoming 1
Incoming 2
Instructions

2. 320
Target X 1
Attempts
Target Y
Civilian Hit 15
WhichSlide 1
Radar Hit
X travel of bomb
181.25 1
Barrier Hit
X Barrier
173
Y Barrier
Ground hit
Trajectory Angle 25
Off
Air
100
Velocity
270
Radar Height
262
Civilian 4 X
254
Civilian 1 X
408
Civilian 5 X
344
Civilian 2 X
373
280
Civilian 3 X -4
Score
Hit Target Flag

3. Click on a link below for instructions
Simple Targets
Barrier
Radar
Radar & Barrier
Elevated Target
Incoming 1
Incoming 2
Calculations
Home

4. Home Score Clear Try New 450 450 50 m/sec Fire 42.5 Instruction Page
Target Position 
X = 296, Y = 0
Home
Score
Find the difference between the target’s x coordinate and the x coordinate for the end of the cannon to determine the distance to the target.
Clear
This will clear all trajectories
Try New
This will re-plot the target and civilians and clear all trajectories.
X coordinate for the end of the cannon.
You’re goal is to get a trajectory that will land the cannon ball on your target. You may try guessing or you can do it mathematically.
Cannon X cor
42.5
Instruction Page
Change angle of the cannon.
These are civilians. You’ll lose points if you hit them. It is also morally reprehensible to target them.
This is your target
The cannon’s elevation is 0.
Air Resistance
Change velocity
From Vert.
From Horiz. 
Off
450
450
50 m/sec
Fire     ?

5. -7 Home Score Clear Try New 450 450 50 m/sec Fire 42.5
Target Position 
X = 530, Y = 0 -7
Home
Barrier Height 
119
Score
Clear
Try New
This is the same as the Simple Target page except now there is a barrier that you must go over in order to hit the target. The barrier is half-way between the target and the cannon. So it is possible to compute the height of your proposed trajectory in order to see if it will clear the barrier. You’ll lose points if you hit the barrier.
Cannon X cor
42.5
Instruction Page
From Vert.
From Horiz. 
Off
450
450
50 m/sec
Fire     ?

6. -9 Home Score Clear Try New 750 150 50 m/sec Fire 58 Instruction Page
Target Position 
X = 446, Y = 0 -9
Home
Radar Height 
192
Score
Clear
Try New
This is the same as the Simple Target page, but now your trajectory may not go above the radar line. You’ll lose points if you do.
Cannon X cor 58
Instruction Page
From Vert.
From Horiz. 
Off
750
150
50 m/sec
Fire     ?

7. Home Score Clear Try New 450 450 50 m/sec Fire 42.5 Instruction Page
Target Position 
X = 550, Y = 0
Home
Radar Height 
195
Score
Clear
Barrier Height 
135
Try New
Things just keep getting more difficult. This is a combination of the Barrier and Radar Pages.
Cannon X cor
42.5
Instruction Page
From Vert.
From Horiz. 
Off
450
450
50 m/sec
Fire     ?

8. 10 Home Score Clear Try New 450 450 50 m/sec Fire Instruction Page
Target Position 
X = 606, Y = 50 10
Home
Radar Height 
229
Score
Clear
Barrier Height 
128
Try New
Now there is an elevated target. Notice the coordinates include the y coordinate above the zero level.
Cannon X cor
Instruction Page
41.5
From Vert.
From Horiz. 
Off
450
450
50 m/sec
Fire     ?

9. Home Score Clear Try New 450 450 50 m/sec Fire 42.5 Instruction Page
Target Position  ?
Home
Radar Height 
267
Score
Barrier Coord. 
X = 252, Y = 132
Clear
Incoming Hit Coord.
Try New
X = 60, Y = 0
Incoming Midpoint
X = 290, Y = 152.5
In this challenge a hidden target has shot at you. The green explosion came from him. The coordinates listed are from the location of the explosion and from the midpoint of the missile’s trajectory.
Cannon X cor
42.5
You must find the distance between the incoming hit and the incoming midpoint. Add this distance to the midpoint coordinate to fine the target coordinate. Subtract the cannon coordinate from the target’s to get the distance from the cannon to the target.
Instruction Page
From Vert.
From Horiz. 
Off
450
450
50 m/sec
Fire     ?

10. Home Score Clear Try New 450 450 50 m/sec Fire 42.5 Instruction Page
Target Position  ?
Home
Radar Height 
248
Score
Barrier Coord. 
X = 301, Y = 175
Clear
Incoming Hit Coord.
Try New
X = 60, Y = 0
Incoming Coord. 1
X = 86, Y = 38
Incoming Coord. 2
X = 218, Y = 172
In this challenge the coordinates of the explosion are given, along with a couple of coordinates the from the second half of the trajectory. This is more of a mathematical challenge.
Cannon X cor
42.5
Instruction Page
From Vert.
From Horiz. 
Off
450
450
50 m/sec
Fire     ?

11. Calculation Help Page Go Back sin(q)vt vt q cos(q)vt v = velocity
Height of trajectory without gravity
v = velocity
t = time (dependent on angle & velocity)
= trajectory angle
a = acceleration of gravity 9.8 m/s/s
Path of trajectory without gravity.
sin(q)vt vt
Height of trajectory including fall due to earth’s gravity. q
sin(q)vt -1/2at2
cos(q)vt
Distance trajectory travels
If we consider a trajectory to start with y = 0 and to end with y = 0, then we can use the equation
y = sin(q)vt – 1/2at2 to predict how long a missile will stay in the air if shot at a given velocity (v) at a given angle to the ground (q). The part of the equation sin(q)vt predicts how high a missile would go after (t) seconds if there was no gravity. The part of the equation 1/2at2 computes the distance an object falls towards earth in (t) seconds. When the distance a missile falls in (t) seconds = the height it would gain after (t) seconds the missile is back on the ground. From this equation we get the following equations.
Maximum Height of a trajectory = sin(q)vt Distance = (cos(q)v)(sin(q)v)
Time in the air = sin(q)v 4
4.9
4.9
Velocity to use at a given angle for a computed distance = (Distance to the target x 4.9)
cos(q) x sin(q)
Go Back

12. 13 2 Home Score Clear Try New 450 450 56 m/sec Fire 42.5
Target Position 
X = 360, Y = 0 13 2
Home
Score
Attempts
Clear
Try New
Cannon X cor
42.5
From Vert.
From Horiz. 
Off
450
450
56 m/sec
Fire     ?

13. 592 1 Home Score Clear Try New 450 450 50 m/sec Fire 42.5
Target Position 
X = 640, Y = 0
592 1
Home
Barrier Height 
Score
Attempts
Clear
Try New
Cannon X cor
42.5
From Vert.
From Horiz. 
Off
450
450
50 m/sec
Fire     ?

14. 529 2 Home Score Clear Try New 450 450 50 m/sec Fire 42.5 Attempts  
Target Position 
X = 420, Y = 0
529 2
Home
Radar Height 
Score
Attempts
Clear
Try New
Cannon X cor
42.5
From Vert.
From Horiz. 
Off
450
450
50 m/sec
Fire     ?

15. -4 1 Home Score Clear Try New 250 650 100 m/sec Fire 32.5 Attempts  
Target Position 
X = 316, Y = 0 -4 1
Home
Radar Height 
270
Score
Attempts
Clear
Barrier Height 
173
Try New
Cannon X cor
32.5
From Vert.
From Horiz. 
Off
250
650
100 m/sec
Fire     ?

16. 88 10 Home Score Clear Try New 390 510 78 m/sec Fire 39.5 Attempts  
Target Position 
X = 560, Y = 100 88 10
Home
Radar Height 
211
Score
Attempts
Clear
Barrier Height 
178
Try New
Cannon X cor
39.5
From Vert.
From Horiz. 
Off
390
510
78 m/sec
Fire     ?

17. -5 3 Home Score Clear Try New 450 450 50 m/sec Fire 42.5 Attempts  
Target Position  ? -5 3
Home
Radar Height 
233
Score
Attempts
Clear
Barrier Coord. 
X = 311, Y = 46
Incoming Hit Coord.
Try New
X = 65, Y = 0
Incoming Midpoint
X = 337.5, Y = 290
Cannon X cor
42.5
From Vert.
From Horiz. 
Off
450
450
50 m/sec
Fire     ?

18. 1 2 Home Score Clear Try New 360 540 33.5 m/sec Fire 38 Attempts   
Target Position  1 2
Home
Radar Height 
290
Score
Attempts
Clear
Barrier Coord. 
X = 240, Y = 100
Incoming Hit Coord.
Try New
X = 70, Y = 0
Incoming Coord. 1
X = 80, Y = 16
Incoming Coord. 2
X = 150, Y = 110
Cannon X cor 38
From Vert.
From Horiz. 
Off
360
540
33.5 m/sec
Fire     ?

20. Congratulations! You finished your trial.88
Your Score is 3
Civilians you hit 6
Times you hit the radar 2
Times you hit the barrier
Times you hit the ground 11
Home

21. Solving Quadratics

22. About Air Resistance Go Back
Air resistance on a moving object is a very complex calculation that involves, the density of the air, the shape of the projectile, the projectile’s surface, and other factors. For this program I made the drag on the projectile a function of the projectile’s speed so that for every meter/sec in velocity there was drag coefficient of Of course as the velocity decreases, so does the total drag. And this amount is multiplied recursively. In other words, the air resistance used in this program did not use any acceptable mathematical model. In still other words, it has no basis in reality; it’s just there for fun.
Go Back