Projectile Motion Example Solved Step by Step for Beginners

 -Dr.Sanjaykumar pawar

Step-by-step projectile motion example showing maximum height, time of flight, and horizontal range.

Internal Links

Laws of Motion Notes

Motion in Two Dimensions

Kinematics Formula Sheet

Newton’s Laws Numerical Problems

Velocity and Acceleration Explained

Circular Motion Notes

Physics Class 11 Important Questions

Projectile Motion Formula Derivation

Work Energy Theorem Notes

Gravitation Chapter Notes

Example 3.8 - Projectile Motion

A cricket ball is thrown at a speed of 28 m/s in a direction 30° above the horizontal.

Calculate:

1. Maximum Height
2. Time Taken to Return to Same Level
3. Horizontal Range

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Step 1: Given Values

Initial velocity (u) = 28 m/s

Angle of projection (θ) = 30°

Acceleration due to gravity (g) = 9.8 m/s²

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(a) Maximum Height

Formula: H = (u² sin²θ) / 2g

H = (28² × sin²30°) / (2 × 9.8)

sin30° = 1/2

H = (28² × (1/2)²) / 19.6

28² = 784

(1/2)² = 1/4

H = (784 × 1/4) / 19.6

H = 196 / 19.6

H = 10 m

Answer: Maximum Height = 10 m

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(b) Time of Flight

Formula: T = (2u sinθ) / g

T = (2 × 28 × sin30°) / 9.8

sin30° = 1/2

T = (2 × 28 × 1/2) / 9.8

T = 28 / 9.8

T = 2.86 s

Approximate value = 2.9 s

Answer: Time of Flight = 2.9 s

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(c) Horizontal Range

Formula: R = (u² sin2θ) / g

R = (28² × sin(2 × 30°)) / 9.8

2 × 30° = 60°

R = (28² × sin60°) / 9.8

sin60° = 0.866

R = (784 × 0.866) / 9.8

R = 678.944 / 9.8

R ≈ 69 m

Answer: Horizontal Range = 69 m

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Final Answers

Quantity	Answer
Maximum Height	10 m
Time of Flight	2.9 s
Horizontal Range	69 m