Projectile Motion
Maximum height reached by the projectile of KN-003 is:
Select the correct option:
Solution
5 m
From KN-003, u=20Ā m/s and Īø=30ā. The formula for maximum height Hmaxā is: Hmaxā=2gu2sin2Īøā Substituting values:
- sin30ā=0.5ā¹sin230ā=0.25
- u2=400
- 2g=20 Hmaxā=20400Ć0.25ā=20100ā=5Ā m.
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About This Question
- Subject
- physics
- Chapter
- kinematics
- Topic
- projectile motion
- Difficulty
- Medium
- Year
- 2025
This medium difficulty physics question is from the chapter kinematics, covering the topic of projectile motion. It appeared in the 2025 exam. Practice this and similar questions to strengthen your understanding of kinematics concepts.
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