Question 15
A stone is thrown vertically upward with an initial velocity of 40 ms-1. Taking g = 10 ms-2, find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?
Given,
Initial velocity (u) = 40 ms-1
Final velocity (v) = 0
g = 10 ms-2
Max height = ?
According to the third equation of motion,
v2 = u2 - 2gs
Note: [negative g as the object goes up]
Substituting we get,
0 = (40)2 - 2 x 10 x s
⇒ 1600 = 20s
⇒ s = = 80 m
Total Distance = sa + sd = 80 + 80 = 160 m
where, sa and sd are distance going up and coming down, respectively.
Displacement = 0 [as the stone comes to the same point from where it was thrown ].
Hence, total distance covered = 160 m and displacement = 0, as the first point is the same as the last point
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)
- 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.
- 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