RealitySim

Mahindra Thar ramp launch at 60 km/h

A Mahindra Thar approaches a 35° incline at 60 km/h and leaves the ramp, following a ballistic trajectory until it lands on level ground.

Inputs & what-if

Approach speed*60 km/h
Incline angle*35 °
Ramp exit height1.20 m
Vehicle mass1850 kg
Drag area (Cd·A)1.60
Gravity9.81 m/s²

Model inspector

Vehicle Ramp Launch (projectile motion)

Vehicle dynamics / 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

Approach speed8.9%
Gravity8.6%
Incline angle7.6%
Ramp exit height0.5%
Drag area (Cd·A)0.0%
Vehicle mass0.0%
0.00 / 2.07 s

Timeline

x

0.00 m

y

1.20 m

|v|

16.67 m/s

a

9.89 m/s²

External data & sources

  • Kerb massreported · medium

    ≈1,850 kg

    Mahindra Thar published specifications · retrieved 1970-01-01

    View source
  • Frontal drag area (Cd·A)estimated · low

    ≈1.6 m²

    Estimated from boxy SUV geometry (Cd ≈ 0.6, A ≈ 2.7 m²) · retrieved 1970-01-01

Assumptions

  • Ramp exit height assumed 1.2 m
  • Level landing surface
  • No traction loss on the ramp
  • 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 time2.07 sconfidence: high

Why did this happen?

  1. Vertical launch component v₀ᵧ = 9.56 m/s
  2. Gravity decelerates then accelerates the body at 9.81 m/s²
  3. Ramp exit height of 1.20 m extends the fall phase
Landing distance28.0 mconfidence: medium

Why did this happen?

  1. Horizontal velocity v₀ₓ = 13.65 m/s
  2. Sustained for the 2.07 s of flight
  3. Reduced slightly by aerodynamic drag; terrain is assumed level
Peak height5.84 mconfidence: high

Why did this happen?

  1. Vertical velocity reaches zero at the apex
  2. Height follows from energy conversion v²/2g
Landing speed17.2 m/sconfidence: medium

Why did this happen?

  1. Horizontal velocity is nearly preserved
  2. Vertical velocity is regained during the fall
Kinetic energy at landing272280 Jconfidence: medium

Why did this happen?

  1. E_k = ½ · 1850 kg · (17.16 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.