• Ballistics •
Ballistics is a part of Physics that deals with the movement
of a body that is thrown into space. In the simplest case, it
is assumed that: (a) only the force of gravity acts; (b) no air
resistance or jet propulsion affects the body; and (c) the surface
of the ground is flat.
We will consider the movement of a gullet fired from a cannon.
The trajectory and movement of the gullet are described by formulas
with three parameters:
- v – initial speed
- g – acceleration of gravity;
- θ – launch angle.
The trajectory is a parabola, with the highest point being reached
mid-flight of the gullet (this only applies if the shot and the fall
of the gullet are at the same level).
• Ballistic cannon •
In this lesson, we will experiment with a virtual ballistic cannon.
Two of the parameters are fixed and one is limited.
- initial speed v = 28 m/s;
- acceleration of gravity g = 9.81 m/s²;
- launch angle θ ∈ [0°,90°].
The general formula for describing the trajectory as a function
of time uses that the horizontal velocity component is constant and
independent of the earth's acceleration:
x(t) = vt.cos(θ)
and
y(t) = vt.sin(θ) – gt²/2
In addition, in the measurements, we will assume that the shot
and hit levels are the same. Therefore, the hit will not be at ground
level, but at the level of the cannon.
Flight duration D(θ) depends only on the launch angle and is calculated in this way:
D(θ) = v².sin(2θ)/g
Flight duration T(θ) is calculates in this way:
T(θ) = 2v.sin(θ)/g
To find the maximum altitude, we used that it is reached mid-flight,
i.e. at t = v.sin(θ)/g.
H(θ) = y[T(θ)/2] = v²sin²(θ)/(2g)
• Field experiment •
To perform a field experiment first determine the direction
of the shot (by moving the mouse cursor over the canvas) and
then click to fire the shot. Once the gullet falls, the
[New Shot] button is used to select a new target.
Until then, the cannon orientation is "locked" and the last
trajectory is visible in the canvas.
The [Measurement] button shows an angle meter,
horizontal and vertical lines with divisions and markers
along the trajectory of the gullet.
The major divisions of the angle gauge are at 10° and the minor
divisions are at 5°. Line divisions are every 10 meters, with intermediate
divisions every 5 meters and small divisions at meter level. The large markers
along the trajectory of the gullet are at 1 second relative to the beginning
of the shot, and the small ones are at half seconds.
New Shot
Measurement
Instructions
• Problems •
In all problems the gullet launches at the same speed of 28 m/s.
When adjusting the initial firing conditions and measuring the results,
use the available ruler and angle gauges.
Show the solution
Problem №1. How far and how high will the gullet fly if it is launched at: (a) 30°, (b) 60° and (c) 90°. Present the results measured by the rulers and compare with the calculated results.
Solution. Switch to "Measurement" mode, adjust the cannon direction
at the appropriate degree and fire a shot. With the vertical line measure the
height of the flight, and with the horizontal line measure the length of the
flight. The results of the measurements and the calculations are shown in the table.
30° |
10 m |
69 m |
 |
? |
? |
60° |
30 m |
69 m |
 |
? |
? |
90° |
40 m |
0 m |
 |
? |
? |
Calculate
Show the solution
Problem №2. The cannon is aimed at 30 meters vertically above a target
that is 50 meters away. Where will the gullet fall: in front of the target, in
the target or behind the target?
Solution. Move the vertical line 50 meters along the horizontal. To properly
target the top, click the 30-th meter on the vertical line. The observed trajectory
indicates that the gullet will fall about 20 meters behind the target.
To verify this we calculate the angle tg(θ) = 30/50, and then we find θ ≈ 30.96°. The length of the flight is calculated with a formula and it is 70.52 meters, i.e. 20.52 meters behind the target. This confirms the measurement.
Show the solution
Problem №3. Determine experimentally at which angle the greatest distance is reached.
Solution. We expect that the distance depends smoothly on the angle, since the trajectory is a parabola. So first we experiment with angles at 10°: 0°, 10°, 20°, ... 90°. The results are shown in the table:
Distance (meters) |
0 |
27 |
51 |
69 |
79 |
79 |
69 |
51 |
27 |
0 |
From the symmetry of the parabolic trajectory and the data in the table,
it can be assumed that the maximum distance is reached at 45°.
Show the solution
Problem №4. Check experimentaly for a given launch angle α∈[5°,40°]
whether: (a) the gullet will fall at the same distance when firing at angles 45°±α;
and (b) whether the flight duration is the same. Justify the result with a formula-based calculation.
Solution. Let's α=10°. Launching at α=35° and α=55°
will make the gullet fall at the same distance of approximately 75 meters:
From the formula for the fight distance and from the properties of cos(x) we get, that at 45°+α and 45°–α the distance is the same:
D(45°–α) = v²sin(90°–2α)/g = v²cos(2α)/g
and
D(45°+α) = v²sin(90°+2α)/g = v²cos(2α)/g
When measuring the flight duration at angles α=35° and α=55°,
we get different times – 3.25 seconds and 5.75 seconds.
Observation shows that a larger angle results in a longer flight.
From the formula for the flight duration and from the monotonic growth
of sin(x) in the angular interval x∈[0°,90°], we get that
for a flight at 45°+α will take longer than a flight at 45°‐α:
T(45°–α) = v.sin(45°–α)/g < v.sin(45°+α)/g = T(45°+α)
because at 45°±α ∈ [0°,90°]:
sin(45°–α) < sin(45°+α)
because:
45°–α < 45°+α