01 · Separation after one second
Find the signed height difference hB − hA at t = 1 s.
Hint
Use 39.2 − 19.6t.
Worked solution
It is 19.6 m: B is still above A.
Understand · explore · practise
Meeting particles in vertical motion
Use a common ground origin and clock to find vertical meetings, check both particles are in flight, and handle delayed launches.
Before you startUpward projections, delayed starts and simultaneous equations.
01 / Use the same origin and clock for both particles
Meeting means equal position at the same overall time.
Write both heights above ground.
Do not equate displacements from different launch points. The initial-height offsets belong in the position equations. Also verify that the meeting occurs after both launches and before either relevant free-flight model ends.
Equal height, different velocitiesExplore
At t = 0, A launches upward from ground at 29.4 m/s and B launches upward from height 39.2 m at 9.8 m/s. Use upward positive, g = 9.8 m/s² and negligible air resistance. Treat them as non-interacting particles so their paths can be compared through the meeting.
02 / Equal accelerations cancel in the height difference
The separation changes linearly while both are in free flight.
A starts at ground with u = 29.4; B starts at 39.2 m with u = 9.8.Worked example
hA = 29.4t − 4.9t²; hB = 39.2 + 9.8t − 4.9t²
Both use the same upward axis and clock.
hB − hA = 39.2 − 19.6t
Their identical quadratic terms cancel.
39.2 − 19.6t = 0 gives t = 2 s
This is the meeting time.
hA(2) = 39.2 m
Check hB(2) gives the same ground height.
Watch: equal height does not mean equal velocity
Pause, replay or seek freely. The notes explain the same idea and stay in view.
02 · Validity at the meeting
Are both particles in free flight at t = 2 s?
Hint
Check their launch times and ground-contact times.
Worked solution
Yes. A returns to ground at 6 s; B reaches ground at 4 s. Two seconds lies after both launches and before either contact.
03 / Calculate each signed velocity separately
A meeting does not require equal speeds or directions.
03 · A at the meeting
Find A’s velocity at t = 2 s.
Hint
vA = 29.4 − 9.8t.
Worked solution
vA = +9.8 m/s, upward.
04 · B at the meeting
Find B’s velocity and direction at the same time.
Hint
vB = 9.8 − 9.8t.
Worked solution
vB = −9.8 m/s, downward. B has passed its apex at 1 s.
05 · Coincidence versus collision
Can you use the equations after contact if the particles actually collide?
Hint
A collision can change their velocities.
Worked solution
Not without a collision model. The explorer explicitly treats the particles as non-interacting. For an actual collision, these equations give the pre-collision event; later motion needs new information.
04 / Replace the delayed particle’s elapsed time everywhere
A common clock does not imply equal flight durations.
A launches from ground at 29.4 m/s; B launches from ground 2 s later with the same upward speed.Worked example
hA = 29.4t − 4.9t²
Overall t begins at A’s launch.
hB = 29.4(t − 2) − 4.9(t − 2)², t ≥ 2
Every time term uses B’s elapsed duration.
hB = 49t − 78.4 − 4.9t²
Expand both the linear and squared terms.
hA = hB gives 19.6t = 78.4, hence t = 4 s
B has then been in flight for 2 s.
Meeting height = 39.2 m
Both are still above ground.
06 · Directions in the delayed case
State both velocities at their t = 4 s meeting.
Hint
A has flown for 4 s; B for 2 s.
Worked solution
vA = 29.4 − 9.8(4) = −9.8 m/s, downward. vB = 29.4 − 9.8(2) = +9.8 m/s, upward.
07 · Mixed-clock error
Why is vB = 29.4 − 9.8(4) wrong at that meeting?
Hint
Four seconds is the overall clock, not B’s flight duration.
Worked solution
B launches at overall t = 2, so use 4 − 2 = 2 s in B’s velocity equation.
05 / Find the elapsed time, then recover the delay
More than one algebraic duration can give a height.
08 · Recover a later launch
A launches from ground at 29.4 m/s. B launches later from ground at the same speed and meets A at overall t = 4 s. Find B’s launch delay.
Hint
At t = 4, A is at 39.2 m. Solve 29.4τ − 4.9τ² = 39.2 for B’s flight duration τ.
Worked solution
τ² − 6τ + 8 = 0 gives τ = 2 or 4. Since B launches later, τ < 4, so τ = 2 and delay d = 4 − 2 = 2 s.
09 · Interpret the rejected branch
What does τ = 4 mean in question 8?
Hint
Calculate d = 4 − τ.
Worked solution
It gives d = 0: simultaneous launches with identical initial conditions. That violates “launches later”.
06 / Equal acceleration is a conditional simplification
Check the model before cancelling terms.
10 · Equal initial speeds, different heights
Two particles start simultaneously with identical vertical velocity but different heights. Can they meet while both have the same constant acceleration?
Hint
Their relative acceleration and initial relative velocity are zero.
Worked solution
No. Their nonzero height difference stays constant while both free-flight assumptions remain valid.
11 · Meeting below ground
An algebraic solution gives a negative height before the stated collision. May you accept it as an airborne meeting?
Hint
Identify first ground contact.
Worked solution
No. The free-flight model has already ended for any particle that reaches the ground earlier; the negative-height intersection is not a valid airborne event.
12 · Different accelerations
Can the quadratic terms be cancelled if air resistance gives different accelerations?
Hint
Compare the actual coefficients.
Worked solution
No. Cancellation requires equal accelerations expressed using the same axis. A different motion model may be needed.
07 / Meeting height is not the distance travelled
A particle may already have passed its apex.
13 · B’s path to the simultaneous meeting
For the first example, find B’s distance travelled before t = 2 s.
Hint
B rises for 1 s and then falls for 1 s.
Worked solution
B rises 4.9 m, from 39.2 to 44.1 m, then falls 4.9 m back to 39.2 m. Distance = 9.8 m, although its net displacement is 0.
14 · A’s path to the same event
Find A’s distance travelled before that meeting.
Hint
A’s apex occurs at 3 s.
Worked solution
A is still ascending at 2 s, so its distance is the rise from ground to 39.2 m: 39.2 m.
08 / Match positions, then interpret each particle’s state
Use shared coordinates and separate elapsed flight times.
Write ground heights with all initial offsets. Use t minus the launch delay where needed. Equate heights, check the shared free-flight domain, and calculate each signed velocity. Split at an apex if asked for distance; the meeting height alone is not a path length.
Section 1 of 8 · Use the same origin and clock for both particles