14
Question Question 14
A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground.
Given,
Initial velocity (u) = 0
Tower height = total distance = 19.6 m
g = 9.8 ms-2
Final velocity (v) = ?
According to the third equation of motion,
v2 = 2gs + u2
Substituting we get,
v2 = (2 x 9.8 x 19.6) + 0
v =
v = 19.6 ms-1
Hence, final velocity = 19.6 ms-1
BRIGHT TUTORIALS
BRIGHT TUTORIALS
CBSE Class IX | Academic Year 2026-2027
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Science | Chapter 9: GravitationWeb Content — Quick Reference
Chapter 9: Gravitation — Quick Reference
Quick Revision Points
- F = Gm₁m₂/r² (G = 6.674 × 10⁻¹¹ N·m²/kg²)
- g = GM/R² = 9.8 m/s² (Earth); g on Moon ≈ g/6
- Mass (kg, constant) vs Weight (N = mg, varies with location)
- Pressure P = F/A (Pa = N/m²); Fluid pressure P = hρg
- Archimedes: buoyant force = weight of displaced fluid
- Float if density < fluid density; Sink if density > fluid density
- Relative density = density of substance / density of water (no unit)
Exam Tips for Chapter 9
- Read the detailed chapter notes for complete coverage of all NCERT topics.
- Practice all NCERT in-text and back exercise questions — they are frequently asked in exams.
- Focus on comparison tables, diagrams, and definitions — these are high-scoring areas.
- For numericals (if applicable), practice at least 20 problems of varying difficulty.
- Refer to the practice question bank (200+ questions) for thorough preparation.
Related Resources
- Detailed Notes: ch09-gravitation.html
- Practice Questions: 100+ questions with answers in 05-practice-questions/
- Chapter Test: 30-mark test paper in 06-tests/chapter-tests-30marks/
- Formula Sheet: Complete formula reference in 03-teacher-aid/formula-sheet.html