Understand · explore · practise
Use given parametric projectile equations to find landing time, horizontal range, maximum height and obstacle clearance. Keep ground height, launch displacement and the valid flight interval distinct.
Before you startQuadratic equations, completing the square and parametric elimination
01 / Read what the vertical coordinate measures
Consider the given model x = 8t, y = 5 + 10t − 5t², where x is horizontal distance in metres, y is height above level ground in metres, and t is seconds after launch. The object starts 5 m above ground.
Landing on this ground means y = 0. Returning to launch height means y = 5.
These are assumed model coefficients. Use the equation given in the question rather than replacing its quadratic coefficient with a remembered value.
x = 8t; y = 5 + 10t − 5t², in metres and seconds.
At t = 0 s, position (0, 5) m; rising.
Use 0 ≤ t ≤ 1 + √2 seconds.
Highest point: (8, 10) m at t = 1 s.
The target height 10 m is reached only at the highest point.
Height y is measured from the ground. Vertical displacement from launch is y − launch height. The model ends at landing; the falling curve is not continued below ground. Axis scales differ.
02 / Find the initial position and interpret the rules
At t = 0: (x, y) = (0, 5)
The horizontal origin is beneath the launch point.
Horizontal distance increases by 8 m each second
The horizontal speed is constant.
Vertical displacement from launch is 10t − 5t²
Subtract the initial height 5.
The quadratic coefficient is negative
The height eventually decreases.
The horizontal speed 8 m/s is not the total speed along the curved trajectory. A horizontal range is likewise not the distance travelled along the arc.
Pause, replay or seek freely. The notes explain the same idea and stay in view.
03 / Solve the ground equation and select the physical root
5 + 10t − 5t² = 0
Use ground height y = 0.
t² − 2t − 1 = 0
Simplify.
t = 1 ± √2
Solve the quadratic exactly.
T = 1 + √2 ≈ 2.414 s
The other root is negative, before the stated launch.
Use the physical interval 0 ≤ t ≤ T. If a different surface has height h, solve y = h instead and decide which meeting corresponds to the event described.
04 / Use the landing time in the horizontal rule
R = 8T = 8(1 + √2)
Substitute the exact positive landing time.
R ≈ 19.31 m
Round only at the end.
If the horizontal rule includes an initial position x₀, the range is x(T) − x₀, not simply the final x-coordinate. The length of the curved flight path is a separate quantity.
05 / Complete the square to locate the highest point
y = 5 + 10t − 5t² = 10 − 5(t − 1)²
Complete the square.
Maximum y = 10 m at t = 1 s
A square is non-negative; this time lies in the flight interval.
x = 8 m
The highest point is (8, 10) m.
Rise above launch = 10 − 5 = 5 m
Maximum ground height and maximum rise differ.
If a stated time interval excludes the vertex, compare permitted endpoints instead. A quadratic vertex outside the physical interval is not an attained maximum for that journey.
06 / Distinguish return height from landing
5 + 10t − 5t² = 5
Set height equal to the initial height.
5t(2 − t) = 0
The solutions are t = 0 and t = 2.
The later return is t = 2 s
It is then at (16, 5) m, still above ground.
At ground-level launch, the initial point already satisfies y = 0. Choose the later positive root for landing, unless the scenario specifies otherwise.
07 / Find both times at a specified height
10 − 5(t − 1)² = 7.5
Use completed-square form.
(t − 1)² = 1/2
Rearrange.
t = 1 ± √2/2
Both values lie inside the flight interval.
x = 8 ± 4√2 m
The earlier meeting is rising; the later one is falling.
A height above 10 m is never reached. Height 10 m is reached once. For a target below the launch height, one quadratic root may be negative, leaving only the descending meeting.
08 / Solve a height condition as an interval
10 − 5(t − 1)² ≥ 8
Start with the condition.
(t − 1)² ≤ 2/5
The inequality reverses when dividing by a negative coefficient.
1 − √10/5 ≤ t ≤ 1 + √10/5
The interval is wholly inside the flight.
Duration = 2√10/5 ≈ 1.265 s
Subtract the entry time from the exit time.
For “strictly above”, exclude the two boundary instants. For a target below ground or after landing, apply the physical time window before interpreting the algebra.
09 / Eliminate time to describe the trajectory
t = x/8
Solve the horizontal rule.
y = 5 + 10(x/8) − 5(x/8)²
Substitute.
y = 5 + 5x/4 − 5x²/64
Simplify to a Cartesian parabola.
0 ≤ x ≤ 8(1 + √2)
Only this part describes the modelled flight.
The Cartesian equation gives the shape, but not the time at each point unless it is kept with the original parametric rule.
10 / Check a wall or target at its horizontal position
8t = 12 gives t = 1.5 s
This lies in the flight interval.
y = 5 + 10(1.5) − 5(1.5)² = 8.75 m
Evaluate height at the wall.
Clearance = 8.75 − 8.5 = 0.25 m
The modelled point clears the top.
This conclusion treats the object as a point. A real object has size, and uncertain measurements or resistance may change the clearance. For a wall with width, check the relevant interval rather than only one horizontal position.
11 / Your turn
x = 6t, y = 12 + 8t − 2t². State the initial position and vertical displacement from launch.
Set t = 0, then subtract its y-value.
Initial position (0, 12); vertical displacement 8t − 2t². Coordinates are in the units specified by the model.
For the preceding rules, find the positive ground-level landing time.
t² − 4t − 6 = 0.
T = 2 + √10; the other root 2 − √10 is negative.
Find the horizontal range for those rules.
Use x(T).
6(2 + √10) coordinate units.
Find the highest point for those rules.
y = 20 − 2(t − 2)².
Maximum height 20 at t = 2, with x = 12. Highest point (12, 20); rise above launch is 8.
When does that flight return to height 12?
8t − 2t² = 0.
t = 0 or 4. The later return is t = 4, before ground landing at 2 + √10.
When is that flight at height 18?
20 − 2(t − 2)² = 18.
t = 1 or 3, giving x = 6 or 18. Both are valid: one rising and one falling.
Can that flight reach height 21?
Compare with the maximum.
No. Its maximum height is 20; the height equation would require a negative squared value.
For how long is the same flight at least height 18?
Use the two boundary times from question 6.
For 1 ≤ t ≤ 3, a duration of 2 time units. Both endpoints are included for “at least”.
Eliminate t from x = 6t, y = 12 + 8t − 2t².
Use t = x/6.
y = 12 + 4x/3 − x²/18. During flight, 0 ≤ x ≤ 6(2 + √10).
Using those rules with metres and seconds, a wall at x = 9 is 19 m high. Does the point clear it?
t = 1.5 s.
Height = 12 + 12 − 4.5 = 19.5 m. It clears by 0.5 m in this point-particle model.
x = 10t, y = 15t − 5t². Find the landing time and range after launch.
5t(3 − t) = 0.
The launch root is 0; later landing is t = 3. The horizontal range is 30 coordinate units.
x = 4 + 6t and y = 10t − 5t². Find the final x-coordinate and horizontal range at landing.
The later ground root is t = 2.
Final x = 16. The range is 16 − 4 = 12, because launch was at x = 4.
For y = 12 + 8t − 2t², when is the height 6 during flight?
Solve t² − 4t − 3 = 0.
t = 2 ± √7. Only 2 + √7 is non-negative; it is before landing 2 + √10 and lies on descent.
The height rule y = 12 + 8t − 2t² is considered only for 0 ≤ t ≤ 1. What maximum is attained there?
The vertex at t = 2 is outside this interval.
The height increases throughout [0, 1], so the maximum occurs at t = 1 and equals 18. The full-flight maximum 20 is not attained in this shorter interval.
An object starts 12 m above ground. A learner finds landing by setting its vertical displacement from launch to zero. What event have they found?
Zero displacement means the original height.
They have found a return to launch height (or the launch instant), not necessarily ground landing. If displacement is s, ground landing requires s = −12 m.
A formula gives negative height after its positive ground root. Should that part automatically be used as the real trajectory?
What changes at ground contact?
No. The model normally ends at the first landing. Rebound, stopping or motion through the surface would need a new justified model.
12 / Recap
Section 1 of 12 · Read what the vertical coordinate measures