01 · Upward axis
A particle is released from rest. With upward positive, state u and a.
Hint
Rest means initial velocity zero.
Worked solution
u = 0 and a = −9.8 m/s².
Understand · explore · practise
Vertical motion: signs and assumptions
Set a consistent axis for vertical motion, distinguish positive g from signed acceleration and state the limits of the constant-gravity model.
Before you startSigned SUVAT quantities, mass and weight.
01 / State the free-flight assumptions
The equations describe motion after release and before contact.
Take gravity as approximately uniform and neglect air resistance.
For these examples use g = 9.8 m/s² as a positive magnitude. Acceleration points downward throughout free flight. With upward positive, a = −g; with downward positive, a = +g. State the value used in your own calculation.
Change the axis, not the fallExplore
A particle is released from rest 30 m above the ground. Take g = 9.8 m/s² and neglect air resistance. Choose an axis and elapsed time; the origin is the release point.
02 / An axis choice determines every sign
Changing the positive direction changes coordinates, not the physical motion.
Watch: one fall, two signed descriptions
Pause, replay or seek freely. The notes explain the same idea and stay in view.
02 · Downward axis
Describe the same release using downward positive.
Hint
The physical acceleration is unchanged.
Worked solution
u = 0 and a = +9.8 m/s². Positive now means downward.
03 / Released from rest does not mean zero acceleration
Its velocity changes immediately after release in the ideal model.
03 · One second after release
Using upward positive, find velocity and displacement after1 s.
Hint
Use u = 0 and a = −9.8.
Worked solution
v = −9.8 m/s and s = −4.9 m. Both point downwards relative to the upward axis.
04 · Same fall, other axis
Give the quantities from question3 with downward positive.
Hint
Reverse signed quantities but keep magnitudes.
Worked solution
v = +9.8 m/s and s = +4.9 m. Speed and distance remain9.8 m/s and4.9 m.
04 / Gravity stays downward during ascent
An upward launch initially moves against the acceleration.
A particle is launched upward at14.7 m/s; use upward positive.Worked example
u = +14.7 m/s, a = −9.8 m/s²
Do not change the sign of acceleration because motion is upward.
At1 s: v = 14.7 − 9.8 = +4.9 m/s
It is still rising but slowing.
At2 s: v = 14.7 − 19.6 = −4.9 m/s
It is now falling; acceleration is still −9.8 m/s².
05 · Turn time
Find the time at the highest point.
Hint
Set v = 0.
Worked solution
t = 14.7/9.8 = 1.5 s.
06 · At the top
State velocity and acceleration at that instant.
Hint
Velocity is zero at a smooth turnaround.
Worked solution
v = 0 but a = −9.8 m/s². The particle does not hover; its downward acceleration continues.
05 / Weight depends on mass; ideal free-fall acceleration does not
Do not confuse the force with the acceleration it produces.
07 · Two weights
Find the weights of masses2 kg and5 kg when g = 9.8 m/s².
Hint
Use W = mg.
Worked solution
The weights are19.6 N and49 N respectively. The heavier body experiences a larger gravitational force.
08 · Two accelerations
Why can both masses have acceleration9.8 m/s² downward when weight is the only force?
Hint
Use F = ma with F = mg.
Worked solution
ma = mg gives a magnitude of g for each nonzero mass. The force increases with mass, while the acceleration is the same in this ideal model.
09 · Air resistance
Would a feather and a compact ball necessarily fall together in air?
Hint
Air resistance may matter differently for the two objects.
Worked solution
No. The equal-acceleration claim assumes negligible air resistance. If drag is significant, that simple free-fall model may not apply.
06 / Choose the origin separately from the direction
A height is a coordinate; displacement is its change.
10 · Height above ground
A particle is dropped from30 m above ground. Find its height after2 s, before impact.
Hint
The downward distance fallen is½gt².
Worked solution
Distance fallen = 19.6 m. Height above ground = 30 − 19.6 = 10.4 m. With upward positive and origin at release, its displacement is−19.6 m.
11 · Initial height
If upward position is measured from the ground, is the initial coordinate zero for question10?
Hint
The origin is now at ground level.
Worked solution
No. Initial position is30 m, while displacement from the starting point is0. Its height law is h = 30 − 4.9t² until impact.
07 / End the free-flight model at contact
Support, drag and impact can change the resultant force.
12 · Supported object
A book rests on a table. Is its acceleration downward at g?
Hint
Weight is not its only force.
Worked solution
No. The supporting force balances its weight and its acceleration is zero. The free-flight assumption is not satisfied.
13 · At impact
May the falling height formula be extended below the ground after the particle hits it?
Hint
Contact introduces a new interaction.
Worked solution
Not as a description of the actual particle. Stop the free-flight model at impact; any rebound or rest requires a new phase.
14 · Constant terminal speed
If a falling object reaches a constant terminal speed in air, is its acceleration still g downward?
Hint
Constant velocity means zero acceleration.
Worked solution
No. Its acceleration is zero when drag balances weight. That situation is outside the model that neglects air resistance.
08 / Keep physical direction and signed quantities consistent
A zero velocity is not proof of zero acceleration.
State the axis, origin, positive g value and free-flight assumptions. Gravity is downward during ascent, at the top and during descent. Weight depends on mass, while ideal free-fall acceleration does not. Check height and stop at contact.
Section 1 of 8 · State the free-flight assumptions