A ball is dropped from the top of a high building with a constant acceleration of 9.8 m/s2. What will be its velocity after 3 seconds?
Correct Answer: Option C
Explanation
1. The question describes a ball dropped from a high building with a constant acceleration of 9.8 m/s² (acceleration due to gravity, g) and asks for its velocity after 3 seconds.
2. This is a problem of linear motion under constant acceleration.
3. The relevant kinematic equation is: v = u + at, where v is final velocity, u is initial velocity, a is acceleration, and t is time.
4. The ball is dropped, so the initial velocity (u) is 0 m/s.
5. The acceleration (a) is given as 9.8 m/s².
6. The time (t) is given as 3 seconds.
7. Substitute the values: v = 0 + (9.8 m/s²) * (3 s).
8. Calculate the final velocity: v = 29.4 m/s.
9. This corresponds to option (C).
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