Estimate how strongly an object rebounds after a vertical drop onto a stationary surface. Enter the drop height and the maximum height of the first rebound in metres. The coefficient of restitution is dimensionless.

\[e = \frac{|v_{\mathrm{after}}|}{|v_{\mathrm{before}}|} = \sqrt{\frac{h_{\mathrm{rebound}}}{h_{\mathrm{drop}}}}\]
  1. Release the object from rest without pushing it or deliberately adding spin.
  2. Measure its vertical drop distance from release to first contact.
  3. Measure the vertical rise from separation to the peak of the first rebound.
  4. Enter both distances in metres.

Use the object’s centre of mass and its position at contact as the reference. For an approximately rigid ball on a horizontal floor, measuring the gap beneath the ball at release and at the rebound peak gives the equivalent distances. Do not mix centre heights above the floor with gaps beneath the ball.

For a drop height of 1.00 m and a rebound height of 0.64 m, e = √(0.64 / 1.00) = 0.80.

The upward speed immediately after impact is 80% of the downward speed immediately before impact. Under the stated assumptions, the vertical translational kinetic energy retained is e² = 0.64, or 64%. The remaining 36% is no longer in that vertical translational motion; it may have transferred into deformation, heat, sound or other motion.

The drop height must be greater than zero. The rebound height must be between zero and the drop height, inclusive. A zero drop height makes the ratio undefined and produces an unavailable result. A rebound above the drop height is outside this calculator’s passive-bounce model and also produces an unavailable result; values are not clamped.

The calculation assumes vertical motion, a stationary horizontal surface, negligible air resistance, constant gravity and approximately equal centre-of-mass heights at contact and separation. Moving surfaces, substantial deformation, spin, sideways motion or added energy can make the height-based estimate unsuitable.

An e of 0 represents no upward rebound; an e of 1 represents equal incident and rebound speed in the vertical direction. The result describes this object–surface interaction under the test conditions, rather than a universal constant for the object.