Question 13
A ball is thrown vertically upwards with a velocity of 49 m/s.
Calculate
(i) The maximum height to which it rises,
(ii) The total time it takes to return to the surface of the earth.
Given,
Initial velocity (u) = 49 ms-1
Final velocity v at maximum height = 0
Acceleration due to earth gravity g = -9.8 ms-2 (negative as ball is thrown up).
Let H be the maximum height to which the ball rises.
By third equation of motion,
2gH = v2 - u2
Substituting we get,
2 × (-9.8) × H = 0 - (49)2
-19.6 H = -2401
H = = 122.5 m
Hence, the maximum height to which it rises = 122.5 m
(ii) Total time T = Time to ascend (Ta) + Time to descend (Td)
According to the first equation of motion,
v = u + gt
Substituting we get,
0 = 49 + (-9.8) x Ta
Ta = = 5 s
Also, Td = 5 s
∴ T = Ta + Td = 5 + 5 = 10 s
Hence, total time it takes to return to the surface of the earth = 10 s
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