Exemplar Problem: Mechanics - Multiple Solution Methods

📋 Problem Statement

Question: A particle is projected vertically upward with initial velocity u. It returns to the ground after time T. Determine the maximum height reached by the particle and the time taken to reach the maximum height.

Given:

• Initial velocity = u (upward)
• Total time of flight = T
• Acceleration due to gravity = g (downward)

To Find:

1. Maximum height (H)
2. Time to reach maximum height (t₁)

🎯 Solution Method 1: Kinematic Equations Approach

Step 1: Analysis of Motion

The motion consists of two phases:

1. Upward motion: From ground to maximum height
2. Downward motion: From maximum height back to ground

Step 2: Using Symmetry of Motion

For a projectile thrown vertically upward:

• Time to go up = Time to come down = T/2
• Initial velocity = Final velocity (in magnitude)
• Acceleration is constant throughout motion

Step 3: Find Maximum Height

For upward motion:

• Initial velocity = u
• Final velocity at top = 0
• Time taken = t₁ = T/2
• Acceleration = -g

Using equation v² = u² + 2as: 0² = u² + 2(-g)H u² = 2gH H = u²/2g

Alternative using equation s = ut + ½at²: H = u(T/2) + ½(-g)(T/2)² H = uT/2 - gT²/8

Since both expressions represent H, we can equate them: u²/2g = uT/2 - gT²/8

Step 4: Find Initial Velocity in Terms of T

From the equation above: u²/2g = uT/2 - gT²/8 Multiply by 8g: 4u² = 4ugT - g²T² g²T² - 4ugT + 4u² = 0 (gT - 2u)² = 0 gT - 2u = 0 u = gT/2

Step 5: Final Results

Maximum Height: H = u²/2g = (gT/2)²/2g = g²T²/4 ÷ 2g = gT²/8

Time to reach maximum height: t₁ = T/2

Answer:

• Maximum Height = gT²/8
• Time to reach maximum height = T/2

🎯 Solution Method 2: Energy Conservation Approach

Step 1: Energy Analysis

Using conservation of mechanical energy:

• At ground: Total energy = KE = ½mu²
• At maximum height: Total energy = PE = mgH
• Total mechanical energy is conserved (ignoring air resistance)

Step 2: Apply Energy Conservation

At ground: E₁ = KE + PE = ½mu² + 0 = ½mu²

At maximum height: E₂ = KE + PE = 0 + mgH = mgH

From conservation of energy: E₁ = E₂ ½mu² = mgH H = u²/2g

Step 3: Use Total Time Information

From Method 1, we know: u = gT/2

Substitute into height equation: H = (gT/2)²/2g = g²T²/4 ÷ 2g = gT²/8

Step 4: Time Analysis

Time to reach maximum height: Using v = u + at: 0 = u - gt₁ t₁ = u/g = (gT/2)/g = T/2

Answer:

• Maximum Height = gT²/8
• Time to reach maximum height = T/2

🎯 Solution Method 3: Graphical Approach

Step 1: Velocity-Time Graph

For vertical motion with constant acceleration:

• Graph is a straight line with slope = -g
• At t = 0: v = u
• At t = t₁: v = 0 (maximum height)
• At t = T: v = -u (returns to ground)

Step 2: Calculate Using Graph Properties

Area under v-t graph = displacement

For complete motion (ground to ground): Total displacement = 0 (returns to starting point)

Graph interpretation:

• Area above x-axis = Area below x-axis
• Area of triangle (0 to T/2) = Area of triangle (T/2 to T)

Area calculation: Area = ½ × base × height = ½ × (T/2) × u = uT/4

Total displacement: Area(upward) - Area(downward) = uT/4 - uT/4 = 0 ✓

Step 3: Find Maximum Height

Maximum height = Area under v-t graph from t = 0 to t = T/2 H = Area = ½ × (T/2) × u = uT/4

But we need to express u in terms of T:

Using velocity equation: v = u - gt At t = T: -u = u - gT 2u = gT u = gT/2

Substitute in height equation: H = (gT/2) × T/4 = gT²/8

Step 4: Time Calculation

Time to reach maximum height = T/2 (from symmetry)

Answer:

• Maximum Height = gT²/8
• Time to reach maximum height = T/2

🎯 Solution Method 4: Integration Approach

Step 1: Set Up Differential Equation

Acceleration: a = dv/dt = -g Velocity: v = dy/dt

Step 2: Integrate to Find Velocity

dv/dt = -g ∫dv = -g∫dt v = -gt + C₁

Initial condition: At t = 0, v = u u = -g(0) + C₁ = C₁

Therefore: v = u - gt

Step 3: Find Time to Reach Maximum Height

At maximum height: v = 0 0 = u - gt₁ t₁ = u/g

Step 4: Integrate to Find Displacement

dy/dt = u - gt ∫dy = ∫(u - gt)dt y = ut - ½gt² + C₂

Initial condition: At t = 0, y = 0 0 = u(0) - ½g(0)² + C₂ = C₂

Therefore: y = ut - ½gt²

Step 5: Apply Boundary Conditions

At t = T: y = 0 (returns to ground) 0 = uT - ½gT² uT = ½gT² u = gT/2

Step 6: Find Maximum Height

Maximum height occurs at t = T/2: H = u(T/2) - ½g(T/2)² H = uT/2 - gT²/8 H = (gT/2)T/2 - gT²/8 H = gT²/4 - gT²/8 H = gT²/8

Answer:

• Maximum Height = gT²/8
• Time to reach maximum height = T/2

🔍 Comparison of Methods

Method 1: Kinematic Equations

• Pros: Quick and straightforward
• Cons: Requires memorization of formulas
• Best for: Students comfortable with standard kinematic formulas

Method 2: Energy Conservation

• Pros: Conceptually elegant, no time calculations needed for height
• Cons: Still requires kinematic analysis for time
• Best for: Students with strong conceptual understanding

Method 3: Graphical Approach

• Pros: Visual understanding, good for intuition
• Cons: Requires graph interpretation skills
• Best for: Visual learners and conceptual understanding

Method 4: Integration

• Pros: Most fundamental approach
• Cons: Mathematically intensive
• Best for: Students with strong calculus background

💡 Key Takeaways

Important Insights:

1. Symmetry: Time up = Time down = T/2
2. Initial velocity: u = gT/2
3. Maximum height: H = gT²/8
4. Multiple approaches: Same problem can be solved using different physics principles

General Problem-Solving Strategy:

1. Identify given information and what needs to be found
2. Choose appropriate method based on your strengths
3. Apply the method systematically
4. Verify results using alternative methods if possible

Exam Tips:

• Choose the method you’re most comfortable with
• Show all steps clearly for full credit
• Check units and reasonableness of answers
• Practice multiple methods for flexibility in exams

🎯 Practice Problems

Similar Problems:

1. A ball thrown upward returns in 4s. Find its initial velocity.
2. A particle reaches maximum height of 20m. Find time taken to reach this height.
3. Using energy method, derive the relationship between H and T.

Challenging Problems:

1. Consider air resistance proportional to velocity. How does it affect the results?
2. A projectile is thrown at angle θ. Find maximum height and range.
3. Derive general formulae for any value of T and u.

Remember: Understanding multiple solution methods makes you a more versatile problem solver. Master different approaches to tackle any question confidently!

Happy Learning! 🚀