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We solve internally in SI, then convert outputs back to your unit system.

Standard problems assume Earth: g ≈ 9.8 m/s².

Height above the landing level. Use h=0 for “already on the ground”.

Positive means downward. Negative means thrown upward.

Air resistance

We use a student-friendly model: Fdrag = k·v. This produces a clean terminal velocity and smooth approach to it.

Options:

Off by default (quiet classroom-safe).

Display:

Chips prefill a scenario and run the calculation.

Result:

No results yet. Enter inputs and click Calculate.

How to use this calculator

  • Enter drop height h and initial velocity v₀ (downward positive).
  • Toggle Air resistance to see terminal velocity and the gauge.
  • Click Calculate for impact time, impact speed, and optional steps + mini viz.

How this calculator works

  • No drag: constant acceleration a=g.
  • With drag: linear drag m·dv/dt = m·g − k·v (downward positive).
  • Terminal velocity: when acceleration → 0, vt = m·g/k.

Formula & Equation Used

Ideal free fall: h = v₀ t + (1/2)g t², v(t)=v₀+g t

Linear drag model: dv/dt = g − (k/m)v

Solutions: v(t)=vt + (v₀−vt)e^{−t/τ}, where τ=m/k

Position: y(t)= (vt)t + (v₀−vt)τ(1−e^{−t/τ}) (drop distance)

Example Problem & Step-by-Step Solution

Example 1 — Ideal drop

  1. Given: h=20 m, v₀=0, g=9.8 m/s²
  2. Solve h = (1/2)g t² → t = √(2h/g)
  3. Impact speed: v = g t

Example 2 — With drag (approaches terminal velocity)

  1. Given: h=60 m, m=2 kg, k=0.8
  2. Compute τ=m/k and vt=m g/k
  3. Solve for t such that y(t)=h, then compute v(t)

Example 3 — Tossed upward, then falls back down (no drag)

  1. Given: h=45 m, v₀=-8 m/s (upward), g=9.8 m/s²
  2. Use h = v₀ t + (1/2) g t² (downward positive).
  3. Solve the quadratic for t and choose the smallest non-negative time (impact).
  4. Compute impact speed with v = v₀ + g t (it will be positive at impact).

Frequently Asked Questions

Q: Why does the “drag ON” speed stop increasing?

Because drag grows with speed. Eventually drag balances weight and acceleration approaches 0 → terminal velocity.

Q: Is this “real” air resistance?

It’s a clean learning model (linear drag). Real drag is often closer to proportional to v², but linear drag is great for intuition and fast solving.

Q: What if I throw the object upward (v₀ negative)?

The calculator still works. It first goes up, then comes back down and hits the ground.

Q: Why do you solve “time to impact” numerically with drag?

Because finding t from y(t)=h isn’t always convenient to invert cleanly in a UI. A fast numeric solve is stable and accurate.