Proposal:
A launcher will be built from a pen housing and a rubber band taped around the front of the tube
and stretched around the back, a skewer/int cartridge will be used as the projectile and loaded from
the from and pulled back and released to launch. The projectile should follow a parabolic arc due to
gravity acting on it during its flight. The time it takes to reach a certain horizontal distance will be
recorded, as well as the angle launched at and the launch height. From this the launch velocity of the
launcher will be determined and the max range and max height will be calculated. The physics
principles that will be used are the relationships between displacement, velocity, and acceleration.
Reporting and design:
Concept:
The launcher will be constructed from a pen housing and a rubber band. The projectile is a stick/ink
cartridge which will act like a dart.
Construction:
A pen housing was used for the main body. A rubber band was cut and taped to the front of the tube
so that it would stretch around the back hole of the tube. A skewer acts as the projectile loaded in
from the front and is pulled back with the rubber band and released to launch.
Testing:
The projectile was tested from a fixed height of 1m and shot at a wall 10m away at an angle of 25°,
the trial was repeated 3 times and the time of flight was recorded using a slow-motion video which
slows the time by a factor of 4. The projectile did not hit anything during its flight until it hit the wall
10m away. It travelled in a parabolic arc.
Calculation and Analysis:
Test
1
Angle from the x axis (°)
25
Start height (m)
1
Flight time (s)
2.68/4 = 0.67
2
3
25
25
1
1
2.70/4 = 0.675
2.66/4 = 0.665
Distance travelled horizontally: 10 m
Average flight time:
(0.67+0.675+0.665)/3 = 0.67s
Max height:
Max vertical displacement of 25° flight if it was not blocked by the wall (assume launch from 0m
high)
1. Find horizontal velocity (d=Vxt)
10 = Vx (0.67)
Vx = 14.925 ms-1
2. Find launch velocity (Vtcos(θ)=Vx)
Vtcos(25) = 14.925 ms-1
Vt = 16.468 ms-1
3. Find vertical velocity (Vtsin (θ)=Vy)
16.468 sin (25) = Vy
Vy = 6.960 ms-1
4. Find how long until vertical velocity = 0. (Vy – 9.8t1 = 0)
6.960 – 9.8 t1 = 0
t1 = 0.71s
5. Find the max vertical displacement ( s = v t1 + 0.5a t12
where a = -9.8 ms-2)
s = 6.960 (0.71) + 0.5 ( - 9.8)( 0.71)2
s = 2.472m
Vertical displacement using flight time to 10m wall: (t = 0.67)
s = 6.960 (0.67) + 0.5 ( - 9.8)( 0.67)2
s = 2.464m
Vertical displacement if 90° angle was used to gain maximum height possible:
Launch speed would be the same (Vt = 16.468 ms-1)
Vt = Vy = 16.468 ms-1
Vy = 0 = 16.468 – 9.8t
t = 1.68s
s = 16.468 (1.68) + 0.5 ( - 9.8)( 1.68)2
s = 13.84m
Max range:
Assuming no air friction. Max range assuming start and end height is the same will be at θ = 45°.
1. Find vertical velocity (Vtsin (θ)=Vy)
16.468 sin (45) = Vy
Vy = 11.644 ms-1
2. Find time of flight ( Vyf = - Vyi
so
Vy – 9.8t = - Vy )
11.644 – 9.8t = -11.644
t = 2.376s
3. Find horizontal velocity (Vtcos (θ)=Vx)
16.468 cos (45) = Vx
Vx = 11.644 ms-1
4. Distance travelled (d=Vxt)
d = 11.644 (2.376)
d = 27.671m
Displacement during test:
Horizontal = 10m (from launcher to target)
Vertical = 2.464m (calculated earlier)
Kinematic:
It accelerates downwards constantly at 9.8 m s-2
The vertical velocity decreases during the length of the flight by 9.8t ms-1
The horizontal velocity does not change during the length of the flight
Dynamic:
Ignoring air friction, the only force the projectile experiences while in motion is the force of gravity
acting in the downwards direction. It travels in a parabolic arc.
Ff
Air friction applies a force opposing the direction of motion
Fg
Energy:
Ignoring air friction, the Mechanical energy of the projectile remains the same during the length of
the flight. While the vertical velocity is positive, the KE is decreasing while the PE is increasing, and
while the vertical velocity is negative, the KE is increasing while the PE is decreasing.
Air friction would reduce the total mechanical energy of the projectile constantly depending on the
velocity of the projectile. It doesn’t change the PE, but does decrease the KE of the projectile
because it reduces velocity and KE = 0.5mv2.