RealitySim

Projectile motion with air drag

A 145 g baseball launched at 30 m/s and 45° from 1.5 m, integrated with quadratic aerodynamic drag.

Inputs & what-if

Launch speed*108 km/h
Launch angle*45 °
Launch height1.50 m
Object mass0.14 kg
Drag area (Cd·A)0.01
Gravity9.81 m/s²

Model inspector

Projectile Motion (point mass)

Classical mechanics

The scenario involves a body leaving a ramp and travelling through the air under gravity, so a ballistic trajectory with optional aerodynamic drag is the appropriate model.

Equations

  • v₀ₓ = v₀·cos(θ), v₀ᵧ = v₀·sin(θ)
  • a = −g ŷ − (½ρ·CdA/m)·|v|·v
  • x(t+Δt) = x(t) + vₓΔt (Δt = 2 ms, semi-implicit Euler)
  • E_k = ½mv², p = mv

Validity range

Speeds 5–200 km/h, angles 0–60°, sub-transonic, near-sea-level air density.

Limitations

  • This is a simplified physics model, not a full multi-body vehicle dynamics simulation.
  • Pitch rotation is illustrative only; real rotation depends on suspension rebound and CG position.
  • Landing damage, rollover and chassis loads are not computed.

Sensitivity

Effect of a +10% change on Air time

Gravity7.2%
Launch angle6.1%
Launch speed5.3%
Drag area (Cd·A)2.5%
Object mass1.8%
Launch height0.4%
0.00 / 3.24 s

Timeline

x

0.00 m

y

1.50 m

|v|

30.00 m/s

a

34.26 m/s²

External data & sources

  • Regulation baseball massofficial · high

    142–149 g

    MLB official rules, ball specification · retrieved 1970-01-01

Assumptions

  • Still air
  • No spin-induced lift (Magnus effect ignored)
  • Vehicle treated as a point mass at its centre of gravity.
  • Ramp exit velocity equals approach speed (no traction/rolling losses on the ramp).
  • Flat, level landing surface at ramp-exit height reference.
  • No suspension travel, tyre deformation or aerodynamic lift.

Results

Air time3.24 sconfidence: high

Why did this happen?

  1. Vertical launch component v₀ᵧ = 21.21 m/s
  2. Gravity decelerates then accelerates the body at 9.81 m/s²
  3. Ramp exit height of 1.50 m extends the fall phase
Landing distance34.6 mconfidence: medium

Why did this happen?

  1. Horizontal velocity v₀ₓ = 21.21 m/s
  2. Sustained for the 3.24 s of flight
  3. Reduced slightly by aerodynamic drag; terrain is assumed level
Peak height13.7 mconfidence: high

Why did this happen?

  1. Vertical velocity reaches zero at the apex
  2. Height follows from energy conversion v²/2g
Landing speed14.3 m/sconfidence: medium

Why did this happen?

  1. Horizontal velocity is nearly preserved
  2. Vertical velocity is regained during the fall
Kinetic energy at landing14.8 Jconfidence: medium

Why did this happen?

  1. E_k = ½ · 0 kg · (14.28 m/s)²

AI explanation assistant

Scientific integrity

Results are computed by an explicit, deterministic model integrated in SI units. Values are labelled as user input, external data, assumption, estimate or calculated result. Simplified models are educational approximations, not engineering-grade predictions.