Unequal-height flights and obstacle clearance
Key idea: For a projectile landing at a different height, solve its vertical motion for flight time. Check each time against whether the projectile is rising or falling before finding range.
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Guided lessons · coming soon
Guided lesson
Guided steps
- Find the event time and keep each physically possible root.
- Use vertical velocity to identify rising and falling events.
- Find both times, then the landing distance and vertical velocity.
- Find the time at the wall before calculating height.
- Calculate height and clearance, then decide whether the ball clears.
- Explain why height at the wall decides clearance.
- Use the landing height when ascent and descent differ.
- Find launch speed, impact velocity and flight time.
- Correct both claims using velocity components and vertical distances.
- Choose the falling event, then find its horizontal distance.
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ContinueA ball kicked from the ground onto a raised platform, or over a wall, does not land at its launch height, so the shortcuts for symmetrical flight no longer work. The method does not change: horizontal and vertical motions, linked by time. What changes is where the time comes from. This lesson shows how to find it, and how to choose between two possible answers.
Where the time comes from
Take the launch point as the origin, with right and up as positive, and neglect air resistance.
| You are given | Find the time from | Then |
|---|---|---|
| the height y at which it lands | y = uyt-(1/2)gt² | find x = uₓt |
| the horizontal distance x to a wall or target | t = x/uₓ | find the height y at that time |
When a target height is above the launch point but below the maximum height, the vertical equation has two positive answers: the projectile passes that height once on the way up and once on the way down. At the maximum height the roots coincide; above it there is no real root and the target cannot be reached. The question must tell you which event you need. When the landing point is below the launch point, one answer is negative and describes a time before launch, so it is rejected.
The symmetry of a level flight also breaks: landing lower than the launch point makes the descent take longer than the ascent.
Landing above the launch point
After solving y = uyt-(1/2)gt², evaluate vy = uy-gt at each root. A positive root means an event after launch; a positive vy means rising, and a negative vy means falling. Choose the root that matches the event described, rather than automatically choosing the smaller positive root.
Guided practice 1
Landing on a higher platform
Problem
A particle is launched with uₓ = 8.00 m s⁻¹ and uy = 12.00 m s⁻¹. It lands, while falling, on a platform 4.00 m above the launch point. Taking g = 9.81 m s⁻², find the time at which it lands and its horizontal distance from the launch point.
Find the landing time and distance
Show the steps Optional working
Show the steps Optional working
Hints
Hint 1
a quadratic
Which vertical-motion equation links the target height to time?
Hint 2
which root?
Check the sign of vy = 12.00-9.81t at each root. Landing while falling needs vy negative.
Show solution step by step
Solve for the times
Reason
The particle passes 4.00 m twice, so there are two roots.Working
Answer
4.905t²-12.00t + 4.00 = 0 gives t = (12.00± square root of (144-78.48))/9.81 = 0.398 s or 2.05 s.
Choose the landing time
Working
Answer
At 0.398 s the particle is still rising; at 2.05 s, vy = 12.00-(9.81)(2.048) = -8.09 m s⁻¹, so it is falling. It lands at 2.05 s, at x = (8.00)(2.048) = 16.4 m.
Clearing a wall
When an obstacle is a known horizontal distance away, the horizontal motion gives the time to reach it. Then work out the height at that time and compare it with the obstacle.
Try it yourself 2
Will it clear the wall?
Problem
A football is kicked at 20.0 m s⁻¹ at 35.0° above the horizontal towards a wall 25.0 m away and 6.00 m high. Neglecting air resistance and taking g = 9.81 m s⁻², does the ball clear the wall, and by how much? Keep unrounded components until the final subtraction and enter the clearance in metres.
Find the height at the wall
Show the steps Optional working
Show the steps Optional working
Hints
Hint 1
time first
Which motion gives the time when the ball reaches the wall?
Show solution step by step
Resolve and find the time
Reason
The constant horizontal velocity sets when the ball reaches the wall.Working
Answer
uₓ = 16.383 m s⁻¹, uy = 11.472 m s⁻¹, so t = 25.0/16.383 = 1.526 s.
Height at the wall
Working
Answer
Using the unrounded components and time, y = (20.0 sin 35.0°)t-(1/2)(9.81)t² = 6.08351… m. The ball clears the wall by 0.0835 m, about 8 cm.
Quick check
The margin is so small that rounding the components early could change the answer. Round only at the end.
Check your understanding 1 Choosing the time at a wall
For the football and wall, explain which motion determines the time and why checking maximum height alone cannot decide clearance. Show your calculation on paper if useful.
Show answer
The wall is at a fixed horizontal coordinate, so t = 25.0/(20.0 cos 35.0°) = 1.52597 s. At that same instant y = (20.0 sin 35.0°)t-(1/2)gt² = 6.08351 m, giving 0.0835 m clearance. The maximum height occurs at a different horizontal position: a ball may rise above the wall’s height elsewhere and still hit it. Compare the shared time, the wall’s coordinate and your final subtraction.
Landing below the launch point
Spot the mistake 3
A ball that just clears a wall at its highest point
Learner claim
A ball is kicked at 50.0° above the horizontal. At its highest point, 8.00 m above the launch point, it just clears a wall. It lands 2.00 m below the launch point. A learner says the ball’s velocity is zero at the wall, and that the descent takes as long as the ascent. Correct both claims, then find the launch speed, the speed and direction at impact, and the total time of flight. Take g = 9.81 m s⁻² and neglect air resistance.
Correct the claims and calculate
Show the steps Optional working
Show the steps Optional working
Show solution step by step
Correct the claim about the top
Method
Only the vertical velocity is zero at the highest point.Reason
Gravity changes only the vertical component.Working
Answer
At the wall vy = 0 but vₓ = u cos 50.0°, so the ball is still moving.
Find the launch speed
Method
Use the 8.00 m rise to find uy.Reason
At the top vy = 0, so 0 = uy²-2gh.Working
Answer
uy = square root of 2(9.81)(8.00) = 12.53 m s⁻¹, so u = 12.53/(sin 50.0°) = 16.4 m s⁻¹.
Find the impact velocity
Reason
From the top, the ball falls 8.00 + 2.00 = 10.0 m starting with zero vertical velocity.Working
Answer
vₓ = 16.36 cos 50.0° = 10.5 m s⁻¹ and vy = - square root of 2(9.81)(10.0) = -14.0 m s⁻¹. The speed is square root of (10.5² + 14.0²) = 17.5 m s⁻¹, at tan⁻¹ 14.0/10.5 = 53.1° below the horizontal.
Correct the symmetry claim
Working
Answer
T = (uy-vy)/g = (12.53 + 14.01)/9.81 = 2.70 s. The ascent takes 12.53/9.81 = 1.28 s and the descent 14.01/9.81 = 1.43 s: the descent is longer because the ball falls further than it rose.
Common mistake
At the highest point only the vertical velocity is zero. Equal ascent and descent times need equal launch and landing heights.
Check your understanding 2 Correct the two claims
For the ball that clears the wall at its highest point then lands lower, correct the zero-velocity and equal-time claims. Explain your impact-velocity and flight-time method.
Show answer
At the highest point only vy is zero; vₓ = u cos 50.0° remains. The 8.00 m rise sets uy = square root of 2g(8.00), then u = uy/sin 50.0°. The descent is 10.0 m, giving vy = - square root of 2g(10.0), which combines with unchanged vₓ for speed and direction. Ascent time is uy/g, descent time is |vy|/g; descent is longer because it covers a greater vertical distance. Compare component directions, both distances and the separate times.
Common mistakes
- Using T = 2uy/g when the landing height differs from the launch height.
- Taking the first root of the quadratic without checking whether the projectile is rising or falling then.
- Finding the time at a wall from the vertical motion instead of from t = x/uₓ.
- Rounding the components early when the answer depends on a small difference.
Before you move on
Check your understanding 3
A projectile has uₓ = 8.0 m s⁻¹ and uy = 9.81 m s⁻¹. Its vertical equation for a raised platform gives times 0.50 s and 1.50 s. It lands on the platform while falling. Take g = 9.81 m s⁻². Use the vertical velocity to choose the landing time and find the platform’s horizontal distance from the launch point.
Show the steps Optional working
Show answer
At 0.50 s, vy = 9.81-9.81(0.50) = +4.905 m s⁻¹, so the projectile is rising. At 1.50 s, vy = -4.905 m s⁻¹, so it is falling and lands then. The horizontal distance is x = uₓt = (8.0)(1.50) = 12 m. Both positive roots describe the same height, but only the later one describes the stated landing event.
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Syllabus and review details
- GCE A-Level H2 Physics 2027 · 2027
Content Overview, PDF pages 9–10; Subject Content, PDF pages 11–30
Last reviewed: