Introduction to Initial Velocity Ratio in Vertical Throw

In physics problems involving a ball thrown vertically up, calculating the v1/v2 ratio helps compare initial velocities for specific heights and speeds. This query—reaching 12 m height with 12 m/s vs maximum height 12 m—tests kinematic mastery for exams like CSIR NET Life Sciences (analytical sections).

Key Physics Concepts

Vertical motion uses v² = u² + 2as where upward is positive, a = -g.

• Case 1 (v₁): At 12 m, v = 12 m/s (not max height).
• Case 2 (v₂): Max height 12 m, v = 0 at top.

Problem Analysis

A man throws a ball vertically upward with initial velocity v₁, reaching a height of 12 m with speed 12 m/s. With v₂, the ball reaches maximum height of 12 m. Use kinematic equations under constant acceleration g = 9.8 m/s² downward.

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Solving for v₂

At maximum height H = 12 m for v₂, final velocity is 0 m/s.

v² = v₂² - 2gH ⟹ 0 = v₂² - 2(9.8)(12) ⟹ v₂² = 235.2 ⟹ v₂ = √235.2 ≈ 15.33 m/s

Correction using precise g=10 m/s²: v₂ = √(2*10*12) = √240 ≈ 15.49 m/s.

Solving for v₁

At height h = 12 m, speed v = 12 m/s (still moving upward).

12² = v₁² - 2g(12) ⟹ 144 = v₁² - 235.2 ⟹ v₁² = 379.2 ⟹ v₁ ≈ 19.48 m/s

With g=10: v₁ = √(144 + 240) = √384 ≈ 19.60 m/s.

Ratio Calculation

v₁/v₂ = √(379.2/235.2) = √1.6122 ≈ 1.27. With g=10: √(384/240) = √1.6 ≈ 1.26 (standard 1.23).

General: v₁/v₂ = √(1 + 6/g), v₂/v₁ = 1 / √(1 + 6/g).

Detailed Step-by-Step Solution

For v₂: 0 = v₂² – 2g(12), so v₂ = √(24g).

For v₁: 12² = v₁² – 2g(12), so v₁ = √(144 + 24g).

Ratio: v₁/v₂ = √(1 + 144/(24g)) = √(1 + 6/g). Using g=10: ≈1.26 (1.23 precise).

Common Mistakes to Avoid

• Assuming 12 m/s is at max height (wrong; v=0 there).
• Ignoring direction (speed is magnitude).
• Wrong g value (use 10 for exam simplicity).

Exam Tips for CSIR NET

Practice similar initial velocity calculations with energy conservation: ½v₁² = ½(12)² + gh.