Vertical Projectile Motion
A ball thrown vertically upward with an initial speed of 20 m/s reaches its highest point, taking g equal to 10 m/s², how long does it stay in the air before returning?
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About This Question
- Subject
- physics
- Chapter
- kinematics
- Topic
- vertical projectile motion
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
4 s
As presented in NCERT Class 11, Chapter 3 (Motion in a Straight Line), a ball thrown upward decelerates at g until its velocity becomes zero at the top, then accelerates downward symmetrically. The time to reach the highest point is t = u/g = 20/10 = 2 s. Because the upward and downward journeys are symmetric for the same height, the total time of flight is twice this, giving 2 × 2 = 4 s. The option 2 s mistakes the ascent time alone for the whole flight. The option 8 s doubles the answer by erroneously using g = 5 m/s². The option 1 s halves the ascent time without justification. Plausibility check: the maximum height reached is u²/(2g) = 400/20 = 20 m, and the time to fall this height from rest is √(2h/g) = √4 = 2 s, which added to the 2 s ascent confirms a total flight time of 4 s.
This medium difficulty physics question is from the chapter kinematics, covering the topic of vertical projectile motion. It appeared in the 2025 exam.
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