01 · Release height
For h=10+15t−5t², find the height at release.
Hint
Release means t=0.
Worked solution
h(0)=10 m above the chosen ground level.
Understand · explore · practise
Domains and validity of mechanics models
Find when a motion formula applies, choose physically relevant roots and distinguish a negative coordinate from an impossible prediction.
Before you startQuadratic equations, completing the square and the mechanics modelling cycle.
01 / An equation does not define its own physical scope
State the event that ends the model.
Algebraic values and physical predictions are different.
A polynomial can be evaluated at any real input. A motion model may apply only after release and before contact. Use the assumptions and the reference level to decide the valid domain.
A flight model ends at contactExplore
Use h=10+15t−5t² metres above level ground. Time t is in seconds after release. This simplified model describes free flight until the first ground contact.
02 / Understand the reference level and initial value
Height is measured from a stated zero.
02 · Later height
Find h at t=2 s.
Hint
Substitute into every term.
Worked solution
h(2)=10+30−20=20 m. The value is above ground and before the first contact.
03 / Solve for the terminating event
Ground contact means h=0 in this coordinate system.
Find when the model reaches level ground.Worked example
10+15t−5t²=0 ⇒ t²−3t−2=0
Divide by −5 and rearrange.
t=(3±√17)/2
The roots are about −0.56155 and 3.56155.
First contact after release: T=(3+√17)/2 s
The negative root is outside this experiment. Use 0≤t≤T for the flight model.
Watch: the valid curve ends at ground contact
Pause, replay or seek freely. The notes explain the same idea and stay in view.
03 · Root choice
Why is the negative root rejected for the flight after release?
Hint
Read the definition of elapsed time.
Worked solution
It corresponds to a time before t=0. It is a root of the extended polynomial, not a ground contact during the stated experiment.
04 · End behaviour
Does this quadratic specify what happens after the object hits the ground?
Hint
The assumptions describe free flight only.
Worked solution
No. Bouncing, deformation or remaining at rest require additional assumptions. Do not extend the flight equation through the contact event.
04 / Check a turning point against the valid interval
A mathematical maximum must also occur during the modelled motion.
Complete the square.Worked example
h=85/4−5(t−3/2)²
The squared term is nonnegative.
Maximum h=85/4=21.25 m at t=1.5 s
This time lies between 0 and T, so the maximum occurs within the flight domain.
05 · Explain the maximum
Why does the completed-square form prove this upper bound?
Hint
A square cannot be negative.
Worked solution
Subtracting 5 times a nonnegative square cannot exceed 85/4. Equality occurs when t−3/2=0.
06 · Restricted interval
If observations only cover 0≤t≤1, does the observed maximum reach 21.25 m?
Hint
The turning point is outside this shorter interval.
Worked solution
No. Height increases on this interval and reaches h(1)=20 m at its endpoint. The full-flight maximum lies later.
05 / Two valid roots can describe ascent and descent
Do not discard a root merely because another one exists.
07 · Height 20 m
Solve h=20 and interpret the two times.
Hint
Rearrange to t²−3t+2=0.
Worked solution
t=1 s or t=2 s. Both lie in the flight domain: the object passes 20 m once while rising and once while falling.
08 · Unreachable height
Can the object reach 22 m in this model?
Hint
Compare with the maximum.
Worked solution
No. The maximum is 21.25 m, so there is no real time with h=22.
06 / A negative coordinate can be meaningful
Its meaning depends on the origin and the physical boundary.
09 · Invalid extension
Calculate h(4) and explain why it is not a free-flight prediction here.
Hint
Contact has already happened before 4 s.
Worked solution
h(4)=10+60−80=−10 m. It is below the level ground, but more importantly t=4 exceeds the first-contact time. The flight model no longer applies.
10 · Different reference
If vertical coordinate y is measured upwards from a balcony, what could y=−3 m mean before landing?
Hint
Zero is the balcony, not necessarily the ground.
Worked solution
The object is 3 m below the balcony. This can be a valid position if it remains above the actual ground or another contact boundary.
11 · Reference change
If the ground coordinate is h and the balcony is 10 m above ground, express y in terms of h.
Hint
Subtract the balcony height.
Worked solution
y=h−10. Ground contact is y=−10, not y=0. A change of origin does not change the physical flight.
07 / Validate the formula as well as the root
A plausible value is not evidence of accuracy by itself.
12 · Simplified gravity
What constant vertical acceleration is represented by the coefficient −5 in this quadratic?
Hint
Compare with h=h₀+ut+(1/2)at², if familiar.
Worked solution
a=−10 m/s² in the upwards-positive convention. This is a simplified constant-acceleration model, not a claim that local gravitational acceleration is exactly 10 everywhere.
13 · Assumption limit
Name one physical effect omitted by this free-flight formula.
Hint
The equation has no varying resistive force.
Worked solution
For example air resistance. A substantial drag force would make constant downward acceleration a poor approximation.
14 · Reporting a domain
State the domain and its interpretation in a complete sentence.
Hint
Include the endpoint and units.
Worked solution
The model applies from release at t=0 until first ground contact at t=(3+√17)/2≈3.562 s, with t measured in seconds.
08 / Keep the physical event beside the algebra
Choose roots using the model, not a memorised rule.
Define the reference level, solve the relevant event equation and keep all roots that fit the physical interval. Check maxima within that interval. Negative coordinates are not automatically impossible; extending a model past the event that ends its assumptions is the real problem.
Section 1 of 8 · An equation does not define its own physical scope