Topic 6.2.212 min read
Paratrooper's Jump
A paratrooper first accelerates toward a high terminal speed. Opening the parachute greatly increases drag, causing deceleration until a much lower terminal speed is reached for safe landing.
Choose Your Level
Beginner Level
A skydiver speeds up after leaving an aircraft. Air resistance grows until it balances weight, giving a fast terminal speed. Opening a parachute creates a much larger area, greatly increases drag and slows the person to a safer terminal speed.
- W
- weight downward (N)
- F_d
- air resistance upward (N)
When to use: Take downward as positive while analysing the fall.
Recommended for Class 11
6.2.2 Paratrooper's Jump
Overview
A paratrooper's motion applies drag, terminal velocity and velocity-time graphs. The textbook graph shows a high terminal speed before the parachute opens and a much smaller terminal speed afterward.
Learning Outcomes
- Describe each stage of a parachute jump.
- Explain why opening a parachute causes deceleration.
- Interpret the textbook velocity-time graph.
- Compare the two terminal speeds.
- Calculate acceleration and displacement from graph data.
- Relate force balance to graph slope.
Intermediate Level
Stage 1: Initial Fall
At t = 0 the paratrooper leaves the aircraft. Speed and drag are initially small, so weight is greater than air resistance. The net force is downward and speed rises rapidly.
Stage 2: First Terminal Speed
As speed increases, drag increases. Eventually drag balances weight. The graph becomes horizontal at approximately 50 m s⁻¹, so the paratrooper falls at a high steady speed.
Stage 3: Parachute Opens
At about t = 42 s the parachute opens. Its large surface area sharply increases air resistance. Drag becomes greater than weight, so the resultant force is upward while the velocity remains downward. This opposite acceleration reduces the downward speed.
Stage 4: Second Terminal Speed and Landing
As the paratrooper slows, drag decreases until it again balances weight. Within about 5 s a new terminal speed near 8 m s⁻¹ is reached. The paratrooper continues at this moderate steady speed and lands at about t = 70 s.
Reading the Textbook Graph
| Graph interval | Shape | Meaning |
|---|---|---|
| 0 to about 25 s | Rising curve with decreasing slope | Speed rises; acceleration gradually decreases |
| About 25 to 42 s | Horizontal near 50 m s⁻¹ | First terminal speed |
| About 42 to 47 s | Steep fall | Parachute opens; strong deceleration |
| About 47 to 70 s | Horizontal near 8 m s⁻¹ | Second terminal speed |
| At about 70 s | Velocity falls to zero | Landing |
Graph Mathematics
The slope of a velocity-time graph gives acceleration. A horizontal part has zero slope and therefore zero acceleration. The area under the graph gives downward displacement.
Solved Numerical Examples
Example 1: Average Deceleration
The textbook graph shows speed falling from 50 m s⁻¹ to 8 m s⁻¹ in about 5 s after the parachute opens. Find the average acceleration, taking downward as positive.
- vi = 50 m s⁻¹
- vf = 8 m s⁻¹
- Δt = 5 s
aavg = vf - viΔt
aavg = 8 - 505
aavg = -425
Final answer: aavg = -8.4 m s⁻², meaning 8.4 m s⁻² upward
Example 2: Distance at Lower Terminal Speed
Using the graph values, estimate the distance fallen from t = 47 s to t = 70 s at 8 m s⁻¹.
- v = 8 m s⁻¹
- Δt = 70 - 47 = 23 s
s = vΔt
s = 8(23)
Final answer: s = 184 m
Example 3: Speed Reduction
Calculate the percentage reduction from the first 50 m s⁻¹ terminal speed to the second 8 m s⁻¹ terminal speed.
- v₁ = 50 m s⁻¹
- v₂ = 8 m s⁻¹
Reduction = v₁ - v₂ = 42 m s⁻¹
Percentage reduction = (4250) × 100
Final answer: Speed is reduced by 84%.
Why the Parachute Works
- The canopy greatly increases effective cross-sectional area.
- Larger area produces much larger drag at a given speed.
- Immediately after opening, drag exceeds weight and slows the fall.
- At lower speed, drag falls until it balances weight again.
- The lower steady speed reduces kinetic energy and landing impact.
Common Mistakes
- Saying the paratrooper moves upward when decelerating.
- Saying force balance means the paratrooper stops.
- Treating the graph's horizontal line as zero velocity.
- Forgetting that graph slope gives acceleration.
- Forgetting that graph area gives displacement.
- Using the Stokes sphere formula directly for a human and parachute.
Short-Question Preparation
Q. Why does a paratrooper initially accelerate?
A. Weight is greater than air resistance, so there is a downward net force.
Q. Why does opening the parachute cause deceleration?
A. The canopy greatly increases area and drag. Drag becomes greater than weight, producing an upward acceleration opposite the downward velocity.
Q. Why is the second terminal speed lower?
A. The open canopy produces the balancing drag at a much smaller speed.
Q. What do horizontal sections of the graph mean?
A. Velocity is constant, graph slope and acceleration are zero, and drag equals weight.
Long-Question Preparation
Q. Explain a paratrooper's jump using forces and a velocity-time graph.
A. 1. At release, drag is small and weight causes downward acceleration. 2. As speed rises, drag increases and acceleration decreases. 3. Drag equals weight at the first terminal speed near 50 m s⁻¹. 4. The graph becomes horizontal. 5. At about 42 s the parachute opens. 6. The larger area makes drag exceed weight. 7. Upward acceleration reduces downward speed. 8. In about 5 s, drag and weight balance again. 9. The second terminal speed is near 8 m s⁻¹. 10. The paratrooper lands at about 70 s. 11. State that slope gives acceleration and area gives displacement.
Sticky Notes
- Before opening: terminal speed ≈ 50 m s⁻¹.
- Parachute opens at ≈ 42 s.
- Drag briefly exceeds weight.
- Deceleration lasts ≈ 5 s.
- After opening: terminal speed ≈ 8 m s⁻¹.
- Landing time ≈ 70 s.
- Slope → acceleration; area → displacement.
60-Second Revision
| Condition | Motion |
|---|---|
| W > F_d | Speeds up downward |
| W = Fd | Terminal velocity |
| F_d > W | Downward motion slows |
Summary
A paratrooper reaches a high terminal speed before opening the parachute. The expanded canopy produces a strong upward drag and deceleration, followed by a lower force-balanced terminal speed suitable for landing.
Ask-the-Tutor Modes
- Explain the four stages simply.
- Help read the graph.
- Explain velocity versus acceleration direction.
- Calculate slope or area.
- Compare the two terminal speeds.
- Test me with an MCQ.
Exam Focus
Exam Focus
Important points
- Four motion stages
- Force comparisons
- Two terminal speeds
- Graph interpretation
- Direction during deceleration
Important formulas
- Fnet = W - Fd
- a = ΔvΔt
- s = area under v-t graph
Common exam traps
- Saying motion reverses
- Horizontal means zero velocity
- Using Stokes sphere law directly
Short question tips
- Quote graph values with about/approximately.
Long question structure
Release → first acceleration → first terminal speed → parachute opens → deceleration → second terminal speed → landing → graph meaning.
Common Mistakes
Common Mistakes
- Mistake
Deceleration does not mean upward motion.
- Mistake
Terminal velocity is non-zero.
- Mistake
Slope gives acceleration.
- Mistake
Area gives displacement.
- Mistake
Do not apply the small-sphere Stokes equation directly to a parachute.
Revision Zone
60-Second Revision
Quick cues from this topic — use them for last-minute revision.
First vt ≈ 50 m s⁻¹.
Open at ≈ 42 s.
Second vt ≈ 8 m s⁻¹.
Land at ≈ 70 s.
Opening a parachute raises drag and creates a much lower terminal speed.
Fnet = W - Fd; a = ΔvΔt
- Definition
- Opening a parachute raises drag and creates a much lower terminal speed.
- Formula
- Fnet = W - Fd; a = ΔvΔt
- Example
- 50 to 8 m s⁻¹ in 5 s gives -8.4 m s⁻² downward-axis acceleration.
- Critical warning
- The paratrooper is still moving downward during deceleration.
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Exam Practice
Important Short Question
Q1. Why does opening a parachute slow a paratrooper?
Conceptual Question
Q2. Does the paratrooper move upward while decelerating?
Important Short Question
Q3. What do slope and area of the graph represent?
Long Question Preparation
Q4. Explain the complete jump using the textbook graph.