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#Free Fall

55 public questions tagged with this topic.

PHYSICS

A stone falls freely from a height of 19.6 m . How long does it take to reach the ground? (Take g = 9.8 m/s² )

Given: A stone falls freely from a height of 19.6 m . How long does it take to reach the ground? (Take g = 9.8 m/s² ) These values define the system as per NCERT data. Formula: For free fall, use y = 1/2 g t². This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Here, y = 19.6 m, g = 9.8 m/s² . Substitute: 19.6 = 1/2 · 9.8 · t² Rightarrow 19.6 = 4.9 t² Rightarrow t² = 4 Rightarrow t = 2 s . The time taken is 2 s . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Units and Measurements and Physical World, Topic: Dimensional analysis and fundamental principles.

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PHYSICS

A stone is dropped from a height of 122.5 m . How long does it take to reach the ground? (Take g = 10 m/s² )

Given: A stone is dropped from a height of 122.5 m . How long does it take to reach the ground? (Take g = 10 m/s² ) Formula: For free fall, use y = 1/2 g t². Substitution & Calculation: Here, y = 122.5 m, g = 10 m/s² . Substitute: 122.5 = 1/2 · 10 · t² Rightarrow 122.5 = 5 t² Rightarrow t² = 24.5 Rightarrow t = √24.5 approx 4.95 s . Rounded to one decimal, the time is 5 s . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Physics Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII, continuing as per NCERT advisory for 2026-27),Topic: Fundamental laws, definitions and applications as per latest NCERT. The section explains governing laws, formulas like μ₀ = 4π.

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PHYSICS

A hot air balloon ascends at 4 m/s for 6 s, then accelerates at 1.5 m/s² for 4 s . A stone is dropped at this instant.

Given: A hot air balloon ascends at 4 m/s for 6 s, then accelerates at 1.5 m/s² for 4 s . A stone is dropped at this instant. How long does it take to reach the ground if the initial height was 30 m ? (Take g = 10 m/s² ) These values define the system as per NCERT data. Formula: Phase 1: h_1 = 4 · 6 = 24 m. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Phase 2: v = 4 + 1.5 · 4 = 10 m/s, h_2 = 4 · 4 + 1/2 · 1.5 · (4)² = 16 + 12 = 28 m . Total height = 30 + 24 + 28 = 82 m, stone’s initial velocity = 10 m/s upward. -82 = 10 t - 5 t² Rightarrow 5 t² - 10 t - 82 = 0 . Solve: t = frac10 pm sqrt100 + 164010 = 10 pm 41.23/10, t = 5.12 s (positive root). Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Units and Measurements and Physical World, Topic: Dimensional analysis and fundamental principles.

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Laws of Motion and Energy Conservation

A 3kg object is dropped from 17m. What is its speed just before hitting the ground? (Take g\=10m/s2)

Potential energy V=mgh=3×10×17=510J. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 18 m/s. This satisfies dimensional consistency and physical conditions given, so option A is scientifically correct.

Ref: NCERT Class 11 Physics > Chapter 6: Work, Energy and Power > Topic: Laws of Motion and Energy Conservation

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Work, Energy and Power - Work-Energy Theorem

A 0.2kg ball is dropped from 12m. What is its speed just before hitting the ground? (Take g\=10m/s2)

Potential energy V=mgh=0.2×10×12=24J. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 14 m/s. This satisfies dimensional consistency and physical conditions given, so option A is scientifically correct.

Ref: NCERT Class 11 Physics > Chapter 6: Work, Energy and Power > Topic: Work, Energy and Power - Work-Energy Theorem

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Kinematic Equations and Uniformly Accelerated Motion

A ball is thrown upwards at 18m/s from a 32m tower. What is its speed when it passes the tower’s base on the way down? (

v2=v02+2gh, v2=(18)2+2⋅10⋅32=324+640=964. As per NCERT Chapter 3, equations v=u+at, s=ut+½at², v²=u²+2as describe uniformly accelerated motion. Applying correct signs and units gives 30 m/s as the result, so option A is correct.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Kinematic Equations and Uniformly Accelerated Motion

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Kinematic Equations and Uniformly Accelerated Motion

Two stones are dropped from a height of 245m, with a 3s interval. What is their separation when the second stone has fal

For free fall, s=½gt² and v²=2gh per NCERT. With g=10 m/s², given height or velocity yields time or distance via these equations. Substituting values gives result matching 165 m. This confirms option C as correct by uniformly accelerated motion equations.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Kinematic Equations and Uniformly Accelerated Motion

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