01 · Direction at each crossing
Find the velocity at the two 39.2 m crossings.
Hint
Use v = 19.6 − 9.8t.
Worked solution
At 1 s, v = +9.8 m/s on ascent; at 3 s, v = −9.8 m/s on descent.
Understand · explore · practise
Times at a height and time above a level
Find both height-crossing times, distinguish a tangent or unreachable level, and restrict time-above intervals to the physical flight.
Before you startUpward projections, quadratics and inequalities.
01 / Intersect the height curve with a horizontal line
One height can occur on ascent and descent.
Solve height = target, then check the flight domain.
For a target below the apex there may be two algebraic times. A root before launch or after contact is not part of the flight. A target at the apex gives one repeated time; a higher target is unreachable.
Time above a horizontal levelExplore
A particle launches upward at 19.6 m/s from height 24.5 m. Its height is h = 44.1 − 4.9(t − 2)² for 0 ≤ t ≤ 5 s, taking g = 9.8 m/s² and negligible air resistance. Choose a target height above ground.
02 / Keep both valid crossing times
Two roots can describe two different events.
When is the model particle 39.2 m above ground?Worked example
24.5 + 19.6t − 4.9t² = 39.2
The target is a ground height, not displacement.
t² − 4t + 3 = (t − 1)(t − 3) = 0
Rearrange after dividing by 4.9.
t = 1 s and t = 3 s
Both lie in the flight domain 0 ≤ t ≤ 5.
Watch: two crossings enclose a time window
Pause, replay or seek freely. The notes explain the same idea and stay in view.
02 · Time strictly above
For how long is the particle strictly above 39.2 m?
Hint
The downward-opening curve is above the level between the roots.
Worked solution
It is above for 1 < t < 3, a duration of 2 s.
03 / Use the shape of the height graph
The interval between crossings is above the target.
03 · At least the target
Give the times when height is at least 39.2 m.
Hint
Include the crossings for ≥.
Worked solution
1 ≤ t ≤ 3. Including the two endpoints does not change the duration of 2 s.
04 · Launch height threshold
Find when the particle is strictly above its 24.5 m launch height.
Hint
Solve 19.6t − 4.9t² > 0.
Worked solution
4.9t(4 − t) > 0 gives 0 < t < 4, lasting 4 s.
05 · A product-sign trap
A learner gives t < 1 or t > 3 for h > 39.2. Explain the sign error.
Hint
Rearranging introduces the factor −4.9.
Worked solution
h − 39.2 = −4.9(t − 1)(t − 3). Thus h > 39.2 requires the product to be negative, which occurs between 1 and 3.
04 / Compare the target with maximum height first
A repeated root is an instant, not a positive duration.
06 · Apex level
When is the model particle at 44.1 m, and for how long is it strictly above that level?
Hint
Use h = 44.1 − 4.9(t − 2)².
Worked solution
It reaches 44.1 m only at t = 2 s. It is never strictly above, so duration is 0 s.
07 · Impossible level
Can it reach 50 m?
Hint
Compare with the maximum 44.1 m.
Worked solution
No. The equation would require (t − 2)² = (44.1 − 50)/4.9 < 0, with no real solution.
05 / A lower level can have only one physical crossing
The particle may already be above the target at launch.
Target height is 19.6 m.Worked example
44.1 − 4.9(t − 2)² = 19.6
Complete-square form makes the roots clear.
t = 2 ± √5
The earlier root is approximately −0.236 s, before launch.
0 ≤ t < 2 + √5
Intersect the above-level interval with the actual flight domain.
Duration = 2 + √5 ≈ 4.24 s
Start timing at launch, not the negative algebraic root.
08 · Above ground
For how long is the particle strictly above ground?
Hint
Its height is positive at launch and first becomes zero at 5 s.
Worked solution
For 0 ≤ t < 5, lasting 5 s. The other algebraic ground-height root, −1 s, is outside the flight.
09 · Why not subtract both roots?
For the 19.6 m target, why is 2√5 s not the flight’s time above the target?
Hint
Part of that algebraic interval predates launch.
Worked solution
The interval from 2 − √5 to 0 is an extrapolation. Only 0 through 2 + √5 belongs to the stated flight.
06 / Use the apex to find the launch speed needed
Required rise is target height minus launch height.
10 · Just reach a target
Find the upward launch speed needed to just reach 44.1 m from a 24.5 m platform.
Hint
At that target v = 0 and the rise is 19.6 m.
Worked solution
u² = 2g(19.6) = 384.16, so u = 19.6 m/s.
11 · A higher target
Find the minimum upward speed needed to reach 49 m from the same platform.
Hint
Required rise is 24.5 m.
Worked solution
u = √(2 × 9.8 × 24.5) = √480.2 ≈ 21.9 m/s. Retain the unrounded value in later calculations.
12 · Just reach versus time above
Does the speed in question 11 give a positive time strictly above 49 m?
Hint
That target is the apex at the minimum speed.
Worked solution
No. It only touches the target height at one instant. A greater upward launch speed is required for a positive time above it.
07 / Report clock times separately from duration
A time window needs a start, an end and a physical interpretation.
13 · Confusing time and duration
A learner says the model is above 39.2 m for 3 s because the later crossing is at 3 s. Correct them.
Hint
The interval begins at 1 s, not launch.
Worked solution
The later crossing is a clock reading. The duration is 3 − 1 = 2 s.
14 · Wrong origin
A learner uses displacement s = 39.2 m for the target height 39.2 m. What should s be?
Hint
Subtract the launch height 24.5 m.
Worked solution
s = 39.2 − 24.5 = 14.7 m. Alternatively write the full ground-height equation including 24.5 m.
08 / Solve crossings, restrict the interval, subtract times
Check reachability and the physical domain.
Use height above the same origin as the target. Find valid crossings, determine which side of the level the curve occupies, and intersect with the flight interval. Subtract physical endpoint times to obtain duration; a repeated root at the apex gives no positive time above the level.
Section 1 of 8 · Intersect the height curve with a horizontal line