01 · Ascent force
For this particle, find the resultant while rising.
Hint
Add the two downward forces.
Worked solution
6 N downward, or −6 N with upward positive.
Understand · explore · practise
Vertical motion with opposing resistance
Build separate ascent and descent force equations when resistance reverses direction, and join the two motion stages at the highest point.
Before you startNewton’s second law, vertical SUVAT and stage clocks.
01 / Resistance opposes motion on each moving branch
Weight stays downward when velocity reverses.
Redraw the force diagram after the highest point.
While rising, both weight and resistance act downward. While falling, weight is downward but resistance is upward. Keeping upward positive, the accelerations are aᵤ = −g − R/m and a_d = −g + R/m. This idealised constant-resistance model is applied separately on the moving branches; it does not specify resistance at the single instant of rest.
Two directions, two force equationsExplore
A 0.5 kg particle is projected upward at 12 m/s from height zero. Assume a constant resistance magnitude R opposing motion on each moving branch, with g = 9.8. All choices satisfy R < mg so the downward branch is possible. The graph ends on return to launch height.
02 / Calculate two accelerations before using SUVAT
Use one sign convention throughout.
A 0.5 kg particle has weight 4.9 N and constant opposing resistance 1.1 N.Worked example
Rising: −4.9 − 1.1 = 0.5aᵤ
Both forces point down.
aᵤ = −12 m/s²
It slows while moving upward.
Falling: −4.9 + 1.1 = 0.5a_d
Resistance now points up.
a_d = −7.6 m/s²
Weight still wins, so downward speed increases.
Watch: resistance changes direction when motion reverses
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02 · Descent force
Find its resultant while falling.
Hint
Resistance now opposes the downward motion.
Worked solution
4.9 − 1.1 = 3.8 N downward, or −3.8 N upward component.
03 / Stop the ascent equation at its first zero velocity
The highest point supplies the initial state for descent.
03 · Time to highest point
The particle starts upward at 12 m/s. Find the ascent time.
Hint
Use aᵤ = −12.
Worked solution
0 = 12 − 12t gives tᵤ = 1 s.
04 · Greatest height
Find the height gained.
Hint
Use v² = u² + 2as on the ascent only.
Worked solution
0 = 144 − 24h gives h = 6 m.
05 · Before the highest point
Find its height after 0.5 s.
Hint
This time lies in the ascent interval.
Worked solution
h = 12(0.5) − 6(0.5)² = 4.5 m.
04 / Restart the clock with zero initial velocity
The descent displacement is negative with upward positive.
06 · Descent duration
Find the time from the highest point until return to launch height.
Hint
Use displacement −6 m, u = 0 and a_d = −7.6.
Worked solution
−6 = ½(−7.6)τ² gives τ = √(12/7.6) = 1.26 s to 3 significant figures. Keep the positive duration.
07 · Return speed
Find the speed when it returns to launch height.
Hint
Use the descent acceleration and displacement.
Worked solution
v² = 2(−7.6)(−6) = 91.2, so velocity is −√91.2 m/s and speed is 9.55 m/s to 3 significant figures.
08 · Total flight time
Find total time until return to launch height.
Hint
Add ascent and descent durations without rounding the latter first.
Worked solution
T = 1 + √(12/7.6) = 2.26 s to 3 significant figures.
09 · During descent
Find height 0.5 s after leaving the highest point.
Hint
Use the descent clock, not the launch clock.
Worked solution
Height above launch = 6 − 3.8(0.5)² = 5.05 m. The overall time is 1.5 s.
05 / Use the correct branch for a specified height
A root must belong to the interval where its equation applies.
10 · Height on ascent
Find the time when it first reaches height 3 m.
Hint
Solve 12t − 6t² = 3, then restrict 0 ≤ t ≤ 1.
Worked solution
t = 1 ± √2/2. The valid ascent time is 1 − √2/2 = 0.293 s to 3 significant figures. The other root is outside the ascent stage.
11 · Height on descent
Find the overall time when it descends through height 3 m.
Hint
Use 6 − 3.8τ² = 3 and then add the 1 s ascent.
Worked solution
τ = √(3/3.8), so overall time is 1 + √(3/3.8) = 1.89 s to 3 significant figures.
06 / Use the acceleration difference to infer resistance
The ascent and descent magnitudes straddle g in this model.
12 · Unknown resistance
A 0.5 kg particle has downward acceleration magnitude 12 m/s² while rising. Find R.
Hint
On ascent, mg + R = 12m.
Worked solution
R = 0.5 × 12 − 4.9 = 1.1 N.
13 · Consistency check
Measured acceleration magnitudes are 12 m/s² on ascent and 7.6 m/s² on descent. Are they consistent with constant opposing resistance and g = 9.8?
Hint
The two magnitudes should average to g when R < mg.
Worked solution
Yes: (12 + 7.6)/2 = 9.8. Also R/m = (12 − 7.6)/2 = 2.2 m/s². For m = 0.5 kg this gives R = 1.1 N.
07 / Do not assume the no-resistance symmetry survives
The upward and downward stages have different accelerations.
14 · A symmetry claim
A learner says descent takes 1 s and return speed is 12 m/s because launch and landing heights match. Explain the error.
Hint
That symmetry requires the same acceleration on both branches.
Worked solution
Here ascent acceleration is −12 but descent acceleration is −7.6 m/s². Descent takes about 1.26 s and return speed is about 9.55 m/s. The no-resistance shortcut does not apply.
This is a simplified model, not a universal drag law. Real resistance can depend on speed. The assumed descent from rest needs R < mg; if a proposed constant opposing resistance is at least mg, do not force this two-branch solution without reconsidering the model.
08 / Join states, not force equations
The highest point separates the two models.
Keep weight downward and reverse resistance with the motion. Find ascent time and height, then start descent with zero velocity and the new acceleration. Use local clocks and reject roots outside their stage.
Section 1 of 8 · Resistance opposes motion on each moving branch