Projectile motion is the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. The object is called a projectile, and its path is called its trajectory.
The program below plots the path of a projectile launced at a certain angle above horizontal and at a certain launch velocity. The program uses the kinematics equations.
For details of kinematics, check the section at the bottom of this page.
The following paragraphs present the background of kinematics.
1.1.1 REVIEW OF KINEMATIC EQUATIONS
As is the custom, initial for velocity and displacement have subscript 0.
x0 = initial position in the x-direction
y0 = the initial position in the y-axis direction
v0 = initial velocity
x, y = position at any time
v = velocity at later time
Assuming constant acceleration, the average velocity is;
ṽ = (1/2)*(v0 + v)
x0
v = v0+ at
x = x0 + v0 + (1/2) at2
v^2 = v0^2 + 2a(x-x0)
Assuming initial position is at the origin, x0 = 0 y0 = 0. The above equations will be simplified as below.
ṽ = (1/2)*(v0 + v)
t
v = v0 + at
x = v0.t+ (1/2) at2
v^2 = v0^2 + 2a.x
Resolving the projectile motion in the horizontal (x-axis) and vertical (y-axis):
Horizontal direction:
ax = 0;
x = vx.t
vx = vox
Vertical direction:
ay = -g = -9.8 m/s^2
vy = v0y – g.t
y = v0y.t – (1/2)g.t^2
vy^2 = v0y^2 – 2.g.y
The final results are obtained by vector addition of displacement vectors x, and y and vectors addition of velocity components vx and vy as follows:
Displacement s = (x^2 + y^2)^(1/2)
Direction of displacement, θ = arctan(y/x)
Velocity, v = (vx^2 +vy^2)^(1/2)
Velocity direction, θv = arctan (vy/vx)
1.1.2 Maximum Height:
The maximum height of the projectile can be found by the eqution:
vy^2 = v0y^2 – 2.g.y
At maximum height, vy = 0
Therefore, 0 = v0y^2 – 2.g.y
Or, y = (v0y^2)/2.g
v0y = v.sinθ
Therefore, the maximum height = (v.sinθ)^2/(2.g) 1)
1.1.3 Time to Maximum Height
The maximum height can be determined by the equation:
y = (1/2) .(v0y +v).t
At maximum height, v = 0,
Therefore, t = 2y/(v0y +v), where y is the maximum height 2)
1.1.4 Horizontal Displacement at maximum Height
For horizontal motion, acceleration ax = 0. There for distance travelled, x = vox.t, where t is the time to maximum height.
Since vx = vcosθ, distance in x-direction = t.vcosθ 3)
1.1.5 Time for projectile to reach ground (y = 0)
The time to reach the maximum height is the same as the time to go from maximum height to ground (launch altitude)
Thus, the total tome for projectile from launch to ground = 2. (Time to maximum height).
Total projectile time = 2.( 2y/(v0y +v) 4)
1.1.6 Total Distance travelled by projectile in the x-direction.
The distance travelled by the projectile in the x-direction is the same as the x-distance travelled from maximum height back to ground.
Therefore, to total distance in the x-direction from launch back to ground x = 2. t.vcosθ
