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Separate free flight from impact, apply a stated rebound-speed rule and calculate the resulting heights, flight times and successive rebounds.
Before you startDrops, upward projections and stage-by-stage SUVAT.
01 / A rebound starts a new free-flight stage
Use the stated rebound rule at contact.
SUVAT with a = −g describes each flight with upward positive. During ground contact an additional force changes the velocity; do not extend one constant-gravity equation across the impact. The speed ratio used here is given as a modelling assumption, not a universal property of bouncing objects.
A particle is dropped from 19.6 m. At each instantaneous ground rebound, its upward speed is half its downward speed immediately before contact. Take g = 9.8 m/s² and ignore air resistance. Select a rebound; heights share a common vertical scale.
02 / Find the velocity just before the first impact
v² = 2gH = 2(9.8)(19.6)
This gives the squared impact speed.
Impact speed = 19.6 m/s
With upward positive, pre-impact velocity is −19.6 m/s.
Drop time = √(2H/g) = 2 s
This is the end of the initial stage.
The rebound upward speed is half the incoming downward speed. State the velocities immediately before and after first contact.
Distinguish speed from signed velocity.
Before contact: −19.6 m/s. After contact: +9.8 m/s.
Can v = u − 9.8t describe that entire velocity change at the bounce?
What force acts during contact?
No. The ground exerts a contact force and the velocity changes from downward to upward. Start a new flight equation using the given post-impact velocity.
03 / Rebound height depends on the square of launch speed
0 = 9.8² − 2(9.8)h
At maximum rebound height the velocity is zero.
h = 4.9 m
This is one quarter of the original 19.6 m height.
If rebound speed is k times impact speed, h = k²H
Substitute u² = k²(2gH) into h = u²/(2g).
Pause, replay or seek freely. The notes explain the same idea and stay in view.
If rebound speed is 0.8 times impact speed, what fraction of the preceding maximum height is reached?
Square the speed ratio.
The height fraction is 0.8² = 0.64.
An object dropped from 20 m rebounds to 5 m. Find the ratio of rebound speed to pre-impact speed in this ideal model.
The height ratio equals the squared speed ratio.
k² = 5/20 = 1/4, so k = 1/2, taking the nonnegative speed ratio.
04 / Reset the local clock after each bounce
How long after the first bounce does the particle reach its apex?
Use 0 = 9.8 − 9.8τ.
τ = 1 s after the bounce, or overall t = 3 s after the initial release.
Find the time from first bounce to second ground contact.
It returns to the same ground level.
Duration = 2(9.8)/9.8 = 2 s. The second ground contact occurs at overall t = 4 s.
Find height 0.5 s after the first bounce.
Use the rebound launch speed 9.8 m/s, not the original drop data.
h = 9.8(0.5) − 4.9(0.5²) = 3.675 m, approximately 3.68 m.
05 / Apply the rule again at the next impact
Find the upward speed immediately after the second ground contact.
The incoming speed is 9.8 m/s.
The new upward speed is 4.9 m/s.
Find its maximum height on that flight.
Use u²/(2g).
h = 4.9²/19.6 = 1.225 m. It is one quarter of the previous rebound height, not one half.
When does the particle next reach ground after the second rebound?
The second rebound flight lasts 2u/g.
That flight lasts 2(4.9)/9.8 = 1 s. Added to the second-contact time 4 s, the third contact is at overall t = 5 s.
06 / Add each separate journey length
Find total distance from initial release to the second ground contact.
Add the initial drop and both legs of the first rebound.
Distance = 19.6 + 2(4.9) = 29.4 m.
Find net displacement from release to the second contact, taking upward positive.
Compare initial and final heights.
Displacement = 0 − 19.6 = −19.6 m. The rebound does not change the endpoint displacement.
07 / Keep collision assumptions separate from flight assumptions
Does the given half-speed rule determine the average force from the ground?
Consider mass and contact duration.
No. The rule specifies velocities, but force calculation also needs further information such as mass and contact duration. Constant-gravity SUVAT is not an impact-force model.
Name two assumptions that may fail for a real bouncing object.
Think about flight and collision separately.
Air resistance may affect flight; contact may take non-negligible time; the rebound-speed ratio may vary between impacts; rotation or deformation may matter. Any two suitable assumptions suffice.
08 / Treat each bounce as a boundary between flights
Find the incoming speed, apply the stated rebound rule, then begin a new upward flight. A speed ratio k gives a height ratio k² for a return to the same ground level with negligible drag. Add flight durations and path lengths separately; do not apply one gravity equation through impact.
Section 1 of 8 · A rebound starts a new free-flight stage