This MCQ module is based on: Free Fall Relative Velocity
TOPIC 7 OF 33
Free Fall Relative Velocity
🎓 Class 11
Physics
CBSE
Theory
Ch 2 – Motion in a Straight Line
⏱ ~14 min
🌐 Language: [gtranslate]
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Free Fall Relative Velocity
Free Fall Under Gravity
Free fall is a special case of uniformly accelerated motion where the only acceleration is due to gravity. Near Earth's surface:
\[g = 9.8 \text{ m/s}^2 \approx 10 \text{ m/s}^2 \text{ (for quick estimates)}\]
Galileo's landmark insight was that all objects, regardless of their mass, fall with the same acceleration when air resistance is negligible. A heavy stone and a light coin dropped from the same height reach the ground at the same time (in vacuum).
Sign Convention for Free Fall
Convention used here: Taking upward as positive:
• Object dropped from rest: \(u = 0\), \(a = -g = -9.8\) m/s²
• Object thrown upward: \(u = +u_0\), \(a = -g\) (decelerates going up, accelerates coming down)
• Object thrown downward: \(u = -u_0\), \(a = -g\)
• Object dropped from rest: \(u = 0\), \(a = -g = -9.8\) m/s²
• Object thrown upward: \(u = +u_0\), \(a = -g\) (decelerates going up, accelerates coming down)
• Object thrown downward: \(u = -u_0\), \(a = -g\)
Worked Example 1 (NCERT Example 2.4): Ball Dropped from Tower
A ball is dropped from the top of a 20 m tower. Find (a) the time to reach the ground, and (b) the velocity just before hitting the ground. (Take \(g = 10\) m/s²)
Solution:
Taking downward as positive: \(u = 0\), \(a = g = 10\) m/s², \(s = 20\) m
(a) Time to reach ground: \[s = ut + \frac{1}{2}gt^2 \Rightarrow 20 = 0 + \frac{1}{2}(10)t^2\] \[20 = 5t^2 \Rightarrow t^2 = 4 \Rightarrow \boxed{t = 2 \text{ s}}\]
(b) Velocity at impact: \[v = u + gt = 0 + 10 \times 2 = \boxed{20 \text{ m/s}}\] Verification: \(v^2 = u^2 + 2gs = 0 + 2(10)(20) = 400 \Rightarrow v = 20\) m/s. Confirmed.
Taking downward as positive: \(u = 0\), \(a = g = 10\) m/s², \(s = 20\) m
(a) Time to reach ground: \[s = ut + \frac{1}{2}gt^2 \Rightarrow 20 = 0 + \frac{1}{2}(10)t^2\] \[20 = 5t^2 \Rightarrow t^2 = 4 \Rightarrow \boxed{t = 2 \text{ s}}\]
(b) Velocity at impact: \[v = u + gt = 0 + 10 \times 2 = \boxed{20 \text{ m/s}}\] Verification: \(v^2 = u^2 + 2gs = 0 + 2(10)(20) = 400 \Rightarrow v = 20\) m/s. Confirmed.
Worked Example 2 (NCERT Example 2.5): Ball Thrown Upward
A ball is thrown vertically upward with a velocity of 25 m/s. Find (a) the maximum height, (b) the total time in the air, and (c) the velocity when it returns to the thrower's hand. (\(g = 10\) m/s²)
Solution:
Taking upward as positive: \(u = +25\) m/s, \(a = -10\) m/s²
(a) Maximum height (at max height, \(v = 0\)): \[v^2 = u^2 + 2as \Rightarrow 0 = 625 + 2(-10)h\] \[20h = 625 \Rightarrow \boxed{h = 31.25 \text{ m}}\]
(b) Time to reach max height: \[v = u + at \Rightarrow 0 = 25 - 10t \Rightarrow t_{\text{up}} = 2.5 \text{ s}\] By symmetry of free fall, time to come down = time to go up. \[\boxed{t_{\text{total}} = 2 \times 2.5 = 5 \text{ s}}\]
(c) Velocity on return:
Using \(v = u + at\) with \(t = 5\) s: \[v = 25 + (-10)(5) = 25 - 50 = \boxed{-25 \text{ m/s}}\] The negative sign indicates the velocity is directed downward. The speed is the same as the initial speed (25 m/s) — energy is conserved.
Taking upward as positive: \(u = +25\) m/s, \(a = -10\) m/s²
(a) Maximum height (at max height, \(v = 0\)): \[v^2 = u^2 + 2as \Rightarrow 0 = 625 + 2(-10)h\] \[20h = 625 \Rightarrow \boxed{h = 31.25 \text{ m}}\]
(b) Time to reach max height: \[v = u + at \Rightarrow 0 = 25 - 10t \Rightarrow t_{\text{up}} = 2.5 \text{ s}\] By symmetry of free fall, time to come down = time to go up. \[\boxed{t_{\text{total}} = 2 \times 2.5 = 5 \text{ s}}\]
(c) Velocity on return:
Using \(v = u + at\) with \(t = 5\) s: \[v = 25 + (-10)(5) = 25 - 50 = \boxed{-25 \text{ m/s}}\] The negative sign indicates the velocity is directed downward. The speed is the same as the initial speed (25 m/s) — energy is conserved.
Stopping Distance
When a vehicle brakes, it decelerates from speed \(u\) to rest (\(v = 0\)). The stopping distance is:
\[d = \frac{u^2}{2|a|}\]
This follows from \(v^2 = u^2 + 2as\) with \(v = 0\): \(0 = u^2 - 2|a|d\), giving \(d = u^2/(2|a|)\).
Critical insight: Stopping distance is proportional to the square of speed. Doubling speed means four times the stopping distance. This is why speed limits near schools and hospitals are strictly enforced.
Worked Example 3 (NCERT Example 2.6): Stopping Distances at Various Speeds
A car can decelerate at 10 m/s². Calculate the stopping distance when travelling at (a) 10 m/s, (b) 20 m/s, (c) 50 m/s.
Solution:
Using \(d = \frac{u^2}{2|a|}\) with \(|a| = 10\) m/s²:
(a) \(d = \frac{10^2}{20} = \frac{100}{20} = \boxed{5 \text{ m}}\)
(b) \(d = \frac{20^2}{20} = \frac{400}{20} = \boxed{20 \text{ m}}\)
(c) \(d = \frac{50^2}{20} = \frac{2500}{20} = \boxed{125 \text{ m}}\)
Pattern: 5 m, 20 m, 125 m. When speed doubles (10→20), distance quadruples (5→20). When speed is 5 times (10→50), distance is 25 times (5→125).
Using \(d = \frac{u^2}{2|a|}\) with \(|a| = 10\) m/s²:
(a) \(d = \frac{10^2}{20} = \frac{100}{20} = \boxed{5 \text{ m}}\)
(b) \(d = \frac{20^2}{20} = \frac{400}{20} = \boxed{20 \text{ m}}\)
(c) \(d = \frac{50^2}{20} = \frac{2500}{20} = \boxed{125 \text{ m}}\)
Pattern: 5 m, 20 m, 125 m. When speed doubles (10→20), distance quadruples (5→20). When speed is 5 times (10→50), distance is 25 times (5→125).
Reaction Time
Reaction time is the delay between perceiving a signal and responding to it. During this time, the vehicle continues at its original speed. The total stopping distance thus has two parts:
\[d_{\text{total}} = d_{\text{reaction}} + d_{\text{braking}} = u \cdot t_r + \frac{u^2}{2|a|}\]
Worked Example 4 (NCERT Example 2.7): Measuring Reaction Time
A ruler is held vertically and dropped. A person catches it after it falls 21.0 cm. Calculate their reaction time. (\(g = 9.8\) m/s²)
Solution:
The ruler falls freely from rest: \(u = 0\), \(s = 0.21\) m, \(a = g = 9.8\) m/s²
\[s = \frac{1}{2}gt_r^2 \Rightarrow 0.21 = \frac{1}{2}(9.8)t_r^2\] \[t_r^2 = \frac{0.42}{9.8} = 0.04286\] \[\boxed{t_r = \sqrt{0.04286} \approx 0.207 \text{ s}}\] This is within the typical range of human reaction times (0.15 to 0.3 s).
The ruler falls freely from rest: \(u = 0\), \(s = 0.21\) m, \(a = g = 9.8\) m/s²
\[s = \frac{1}{2}gt_r^2 \Rightarrow 0.21 = \frac{1}{2}(9.8)t_r^2\] \[t_r^2 = \frac{0.42}{9.8} = 0.04286\] \[\boxed{t_r = \sqrt{0.04286} \approx 0.207 \text{ s}}\] This is within the typical range of human reaction times (0.15 to 0.3 s).
2.7 Relative Velocity
The velocity of an object is always measured relative to some reference frame. The relative velocity of object A with respect to object B is:
\[v_{AB} = v_A - v_B\]
Two Objects Moving in the Same Direction
If A moves at \(v_A\) and B at \(v_B\) in the same direction, A appears to move at \(v_A - v_B\) relative to B. If \(v_A > v_B\), A moves forward relative to B.
Two Objects Moving in Opposite Directions
If A moves at \(v_A\) (positive direction) and B at \(v_B\) (negative direction), then \(v_B = -|v_B|\):
\[v_{AB} = v_A - (-|v_B|) = v_A + |v_B|\]
They approach each other with a relative speed equal to the sum of their individual speeds.
Worked Example 5: Two Trains Approaching
Two trains are 200 km apart on the same track, heading towards each other. Train A travels at 60 km/h and Train B at 40 km/h. How long before they meet?
Solution:
Since the trains move toward each other, the relative speed is: \[v_{\text{rel}} = 60 + 40 = 100 \text{ km/h}\] Time to close a 200 km gap: \[t = \frac{\text{distance}}{v_{\text{rel}}} = \frac{200}{100} = \boxed{2 \text{ hours}}\] Verification: In 2 hours, A covers \(60 \times 2 = 120\) km and B covers \(40 \times 2 = 80\) km. Total = 200 km. Confirmed.
Since the trains move toward each other, the relative speed is: \[v_{\text{rel}} = 60 + 40 = 100 \text{ km/h}\] Time to close a 200 km gap: \[t = \frac{\text{distance}}{v_{\text{rel}}} = \frac{200}{100} = \boxed{2 \text{ hours}}\] Verification: In 2 hours, A covers \(60 \times 2 = 120\) km and B covers \(40 \times 2 = 80\) km. Total = 200 km. Confirmed.
Worked Example 6: Overtaking Problem
A car travelling at 90 km/h wants to overtake a truck 20 m ahead moving at 72 km/h. How long does it take the car to reach the truck?
Solution:
Converting to m/s: car speed = \(90 \times \frac{5}{18} = 25\) m/s, truck speed = \(72 \times \frac{5}{18} = 20\) m/s
Relative velocity of car w.r.t. truck = \(25 - 20 = 5\) m/s
Time to cover 20 m gap: \[t = \frac{20}{5} = \boxed{4 \text{ s}}\]
Converting to m/s: car speed = \(90 \times \frac{5}{18} = 25\) m/s, truck speed = \(72 \times \frac{5}{18} = 20\) m/s
Relative velocity of car w.r.t. truck = \(25 - 20 = 5\) m/s
Time to cover 20 m gap: \[t = \frac{20}{5} = \boxed{4 \text{ s}}\]
Activity: Ruler Drop — Measure Your Reaction Time
L3 Apply
Predict: Do you think your reaction time is more or less than a quarter of a second? How does it compare between your dominant and non-dominant hand?
- Have a partner hold a 30 cm ruler vertically with the 0 cm mark at the bottom, aligned with the top of your open thumb and index finger.
- Without warning, your partner drops the ruler. Catch it as quickly as possible.
- Record the distance \(d\) (in metres) at which you caught it.
- Calculate reaction time: \(t_r = \sqrt{\frac{2d}{g}}\) with \(g = 9.8\) m/s².
- Repeat 5 times and find the average. Try with each hand.
Typical values:
If \(d = 15\) cm = 0.15 m: \(t_r = \sqrt{0.30/9.8} = \sqrt{0.0306} \approx 0.175\) s
If \(d = 25\) cm = 0.25 m: \(t_r = \sqrt{0.50/9.8} = \sqrt{0.051} \approx 0.226\) s
Most people have reaction times between 0.15 s and 0.3 s. The dominant hand is usually slightly faster. Fatigue and distractions increase reaction time — which is why using a phone while driving is so dangerous.
If \(d = 15\) cm = 0.15 m: \(t_r = \sqrt{0.30/9.8} = \sqrt{0.0306} \approx 0.175\) s
If \(d = 25\) cm = 0.25 m: \(t_r = \sqrt{0.50/9.8} = \sqrt{0.051} \approx 0.226\) s
Most people have reaction times between 0.15 s and 0.3 s. The dominant hand is usually slightly faster. Fatigue and distractions increase reaction time — which is why using a phone while driving is so dangerous.
Interactive: Free Fall Simulator L3 Apply
Set the drop height and watch the ball fall. Time and velocity are calculated in real-time.
Competency-Based Questions
A highway safety study measures the stopping behaviour of cars at different speeds. The road has a friction coefficient that allows a maximum deceleration of 8 m/s². The average driver reaction time is 0.6 seconds.
Q1. L1 Remember What is the formula for stopping distance in terms of initial speed \(u\) and deceleration \(a\)?
Answer: B. \(d = u^2/(2|a|)\). This comes from setting \(v = 0\) in \(v^2 = u^2 + 2as\).
Q2. L3 Apply Calculate the total stopping distance (reaction + braking) for a car at 72 km/h. (3 marks)
Answer:
\(u = 72 \text{ km/h} = 20\) m/s
Reaction distance = \(u \times t_r = 20 \times 0.6 = 12\) m
Braking distance = \(u^2/(2 \times 8) = 400/16 = 25\) m
Total = \(12 + 25 = \boxed{37 \text{ m}}\)
\(u = 72 \text{ km/h} = 20\) m/s
Reaction distance = \(u \times t_r = 20 \times 0.6 = 12\) m
Braking distance = \(u^2/(2 \times 8) = 400/16 = 25\) m
Total = \(12 + 25 = \boxed{37 \text{ m}}\)
Q3. L4 Analyse A stone is dropped from a cliff. It covers 35% of its total fall distance in the last second. Find the total height of the cliff. (\(g = 10\) m/s²) (4 marks)
Answer:
Let total time of fall = \(T\) seconds. Total height \(H = \frac{1}{2}gT^2 = 5T^2\).
Distance in last second = distance in \(T\) seconds minus distance in \((T-1)\) seconds:
\[d_{\text{last}} = 5T^2 - 5(T-1)^2 = 5[T^2 - T^2 + 2T - 1] = 5(2T-1)\] Given: \(d_{\text{last}} = 0.35H = 0.35 \times 5T^2 = 1.75T^2\)
\[5(2T-1) = 1.75T^2\] \[10T - 5 = 1.75T^2\] \[1.75T^2 - 10T + 5 = 0 \Rightarrow 7T^2 - 40T + 20 = 0\] \[T = \frac{40 \pm \sqrt{1600 - 560}}{14} = \frac{40 \pm \sqrt{1040}}{14} = \frac{40 \pm 32.25}{14}\] \(T = 5.16\) s or \(T = 0.554\) s. Since the last second condition requires \(T > 1\), we take \(T \approx 5.16\) s.
\[H = 5(5.16)^2 \approx \boxed{133 \text{ m}}\]
Let total time of fall = \(T\) seconds. Total height \(H = \frac{1}{2}gT^2 = 5T^2\).
Distance in last second = distance in \(T\) seconds minus distance in \((T-1)\) seconds:
\[d_{\text{last}} = 5T^2 - 5(T-1)^2 = 5[T^2 - T^2 + 2T - 1] = 5(2T-1)\] Given: \(d_{\text{last}} = 0.35H = 0.35 \times 5T^2 = 1.75T^2\)
\[5(2T-1) = 1.75T^2\] \[10T - 5 = 1.75T^2\] \[1.75T^2 - 10T + 5 = 0 \Rightarrow 7T^2 - 40T + 20 = 0\] \[T = \frac{40 \pm \sqrt{1600 - 560}}{14} = \frac{40 \pm \sqrt{1040}}{14} = \frac{40 \pm 32.25}{14}\] \(T = 5.16\) s or \(T = 0.554\) s. Since the last second condition requires \(T > 1\), we take \(T \approx 5.16\) s.
\[H = 5(5.16)^2 \approx \boxed{133 \text{ m}}\]
Q4. L3 Apply Two buses start from towns 300 km apart. Bus P travels at 40 km/h and Bus Q at 60 km/h, both heading toward each other. After how long do they meet, and how far from P's starting point? (3 marks)
Answer:
Relative speed = 40 + 60 = 100 km/h (approaching each other).
Time to meet = 300/100 = 3 hours.
Distance from P's start = \(40 \times 3 = \boxed{120 \text{ km}}\)
Relative speed = 40 + 60 = 100 km/h (approaching each other).
Time to meet = 300/100 = 3 hours.
Distance from P's start = \(40 \times 3 = \boxed{120 \text{ km}}\)
Q5. L5 Evaluate Rain falls vertically at 10 m/s. A person walks at 5 m/s. At what angle should they hold an umbrella to stay dry? Would running faster change this angle? (3 marks)
Answer:
Relative to the person, rain has a horizontal component of 5 m/s (backward) and vertical component of 10 m/s (downward).
Angle with vertical: \(\theta = \tan^{-1}\left(\frac{5}{10}\right) = \tan^{-1}(0.5) \approx \boxed{26.6°}\)
The umbrella should be tilted 26.6° from vertical in the direction of walking.
If they run faster (say 10 m/s): \(\theta = \tan^{-1}(10/10) = 45°\). Running faster increases the angle. The faster you move, the more tilted the umbrella must be. At very high speeds, the rain would appear almost horizontal.
Relative to the person, rain has a horizontal component of 5 m/s (backward) and vertical component of 10 m/s (downward).
Angle with vertical: \(\theta = \tan^{-1}\left(\frac{5}{10}\right) = \tan^{-1}(0.5) \approx \boxed{26.6°}\)
The umbrella should be tilted 26.6° from vertical in the direction of walking.
If they run faster (say 10 m/s): \(\theta = \tan^{-1}(10/10) = 45°\). Running faster increases the angle. The faster you move, the more tilted the umbrella must be. At very high speeds, the rain would appear almost horizontal.
Assertion-Reason Questions
Assertion (A): A ball thrown upward has zero velocity but non-zero acceleration at its highest point.
Reason (R): The acceleration due to gravity acts downward at all times during the flight, regardless of the direction of velocity.
Answer: A. At the highest point, the ball momentarily stops (v = 0) but is still being pulled downward by gravity (a = −g ≠ 0). The reason correctly explains this: gravity acts continuously downward, independent of the ball's motion.
Assertion (A): The stopping distance of a car doubles when its speed doubles.
Reason (R): Stopping distance is proportional to the square of the initial speed.
Answer: C. The assertion is false — when speed doubles, stopping distance quadruples (not doubles). The reason is true: \(d = u^2/(2|a|)\), so \(d \propto u^2\). Since the reason is correct, it actually disproves the assertion.
Assertion (A): When two cars move in the same direction, the relative velocity equals the sum of their speeds.
Reason (R): Relative velocity of A with respect to B is \(v_{AB} = v_A - v_B\).
Answer: D. A is false: for the same direction, relative speed = difference (not sum) of speeds. R is true: \(v_{AB} = v_A - v_B\) is the correct formula. Since they move in the same direction, both velocities have the same sign, so the relative speed is their difference.
Frequently Asked Questions - Free Fall Relative Velocity
What is the main concept covered in Free Fall Relative Velocity?
In NCERT Class 11 Physics Chapter 2 (Motion in a Straight Line), "Free Fall Relative Velocity" covers core principles and equations needed for board exam success. The MyAiSchool lesson explains the topic with definitions, derivations, worked examples, and interactive simulations. Key formulas and dimensional analysis are included to build conceptual depth and problem-solving skills aligned with the CBSE 2025-26 syllabus.
How is Free Fall Relative Velocity useful in real-life applications?
Real-life applications of Free Fall Relative Velocity from NCERT Class 11 Physics Chapter 2 include engineering design, satellite mechanics, sports biomechanics, transportation safety, and electrical/electronic devices. The MyAiSchool lesson links every concept to a tangible example so students see physics as a problem-solving framework for the physical world, not as abstract formulas.
What are the key formulas in Free Fall Relative Velocity?
Key formulas in Free Fall Relative Velocity (NCERT Class 11 Physics Chapter 2 Motion in a Straight Line) are derived step-by-step in the MyAiSchool lesson. Students should memorize the final formula AND understand its derivation for full board marks. Each formula is listed with its dimensional formula, SI unit, applicability range, and common pitfalls. The Summary section at the end of each part includes a quick-reference formula card.
How does this part connect to other parts of Chapter 2?
NCERT Class 11 Physics Chapter 2 (Motion in a Straight Line) is structured so each part builds on the previous one. Free Fall Relative Velocity connects directly to neighbouring parts via shared definitions, units, and methodology. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Free Fall Relative Velocity?
CBSE board questions from Free Fall Relative Velocity typically include: (1) 1-mark MCQs on definitions and formulas, (2) 2-mark short-answer derivations or applications, (3) 3-mark numerical problems with units, (4) 5-mark long-answer derivations followed by application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the Free Fall Relative Velocity lesson allows students to adjust input parameters (sliders or selectors) and see physical quantities update in real time. To use it effectively: (1) try extreme values to understand limiting cases, (2) compare with the analytical formula, (3) check unit consistency, (4) test special configurations from worked examples. The simulation reinforces conceptual intuition that pure formula manipulation cannot.
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