Projectile Motion – Simulation Introduction

Projectile motion describes the two-dimensional trajectory of an object launched with initial speed _V₀ at some angle to the horizontal.

This motion can be viewed as the combination of two independent components: one horizontal and one vertical. The horizontal component remains constant throughout the flight, while the vertical component decreases by g·t due to gravity pulling the object downward (free fall). At Earth’s surface, the gravitational acceleration is approximately g = 9.81 m/s².

Applications of Projectile Motion

Projectile motion isn’t just a textbook example in kinematics—it plays a key role in many everyday activities and sports. Here are a few highlights that show why understanding the trajectory of a launched object matters:

• Ballistics and safety: In designing firearms or fire-hose nozzles, engineers must predict exactly where a stream of water or projectile will land to maximize effectiveness and minimize risk.
• Engineering and irrigation: When creating water fountains or irrigation systems, the launch angle of water jets determines how far the spray will reach the fields, ensuring even coverage.
• Theme parks: The design of water slides and rides often relies on projectile principles so that riders descend safely into the splash pool at just the right spot.

In each case, applying the basic equations for horizontal and vertical motion allows for accurate trajectory predictions, optimal launch angles, and precise range control—whether you’re scoring a point in a game or watering crops with pinpoint accuracy.

Key Equations for Projectile Motion

Horizontal velocity component:
v_0x = v₀ · cos α

Vertical velocity component:
v_0y = v₀ · sin α − g·t

Horizontal displacement:
x = v₀ · t · cos α

Vertical displacement:
y = v₀ · t · sin α − ½·g·t²

Time to reach maximum height:
t_top = (v₀ · sin α) / g

Maximum height (above launch point):
H_max = (v₀² · sin² α) / (2·g)

Range (from same launch height):
D = (v₀² · sin 2α) / g

Projectile Launched from Initial Height H₀

Consider the simulation shown in Figure 1:

Here a projectile starts at height H0 above ground with initial speed v0 at angle θ above the horizontal. Try the interactive version on the Simulation page.

Determining Range and Maximum Height

1. Total flight time
tD

The horizontal range is

D = v0x · tD

where v0x = v0 cos θ.
The total flight time

tD = 2·t1 + tP

– t1 is the time to rise from H0 to the apex (and the same time to descend back to H0),
– t_P is the time to fall from H0 to the ground.

2. Ascent and descent to the same height

From vertical motion with initial upward speed v0y:

0 = v0y − g·t₁  ⇒  t₁ = (v0y) / g

3. Fall time from height H0

Falling from H0 with initial downward speed v0y solves

½·g·tP² + v0y·tP − H₀ = 0

The positive root is:

tP = (−v0y + √v0y² + 2·g·H0) / g

4. Final flight time

Substituting gives

      tD = 2·(v0y/g) + (−v0y + √v0y² + 2·g·H0)/g 
 
      tD= (v0y + √v0y² + 2·g·H0) / g
    

5. Range

D = v0x · tD

6. Maximum height above ground

The rise above H0 is

ΔH = (v0y²) / (2·g)

Hence total maximum height:

Hmax = H0 + (v0y²) / (2·g)