Projectile analysis almost always starts with one simple question: how much of the velocity is horizontal, and how much is vertical? If the initial speed is v0 and the launch angle is θ, the initial components are:
v0x=v0cosθ
v0y=v0sinθ
After that, the horizontal and vertical directions are calculated with different rules. The horizontal direction uses uniform motion because there is no horizontal acceleration. The vertical direction uses uniformly accelerated motion because gravity keeps acting downward.
The key to projectile analysis is using the same time for two different directions.
Cannonball Component Analysis
A cannonball crosses the water while the ghost balls mark equal time steps from the same formulas used in the calculation.
Horizontal component
v0x=20 m/s
Vertical component
v0y=34.6 m/s
Peak time
t=3.5 s
Flight time
T=6.9 s
Range
R=138.6 m
Velocity at two seconds
v=⟨20,14.6⟩ m/s
The arc and ghost balls show positions at equal time intervals, not separate horizontal and vertical paths. The values below the scene come from the same split: v0x stays constant, while vy=v0y−gt changes because gravity acts downward.
The number flow is easier to read as one connected formula sequence:
So the values below the scene are not manually chosen numbers. They are calculated from the selected launch, then shown as components, time, range, and instantaneous velocity.
At the peak, the object is still moving horizontally, but its up-and-down motion pauses for an instant. So the peak condition is vy=0.
vy=v0y−gt
When vy=0, the time to the peak is:
tpeak=gv0y
For an initial speed of 40 m/s at an angle of 60∘, the initial vertical component is 203 m/s. Before the peak it is positive, at the peak it becomes zero, and after the peak it is negative because the object is moving downward.
To find velocity at a specific time, calculate vx and vy at that time first. The horizontal component stays constant, while the vertical component changes.
vx=v0x
vy=v0y−gt
v=vx2+vy2
The direction of velocity can also be read from the ratio of the vertical and horizontal components:
tanα=vxvy
If vy is positive, the velocity points upward. If vy is negative, the velocity points downward.
A ball is launched with an initial speed of 40 m/s and an angle of 60∘. Use g=10 m/s2.
This example uses an angle with clear trigonometric components, so the full analysis stays visible: split the initial velocity, get time from vertical motion, then use the same time to read range and instantaneous velocity.
The initial components are:
v0x=40cos60∘=20 m/s
v0y=40sin60∘=203 m/s
The time to the peak is:
tpeak=10203=23 s
The total time and range are:
ttotal=43 s
R=20(43)=803 m
Suppose we want the velocity at t=2 s. The vertical component is:
vy=203−10(2)=203−20 m/s
Meanwhile, vx=20 m/s stays constant. The speed is:
v=202+(203−20)2=2000−8003≈24.8 m/s
The direction of the velocity from the horizontal is:
α=tan−1(20203−20)≈36.2∘
So at t=2 s, the ball is still moving upward at an angle because vy is still positive.