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 ?

For an object released from rest under constant gravitational acceleration, use the kinematic relation v = v0 + a t to find the speed after time t; here v0 is the initial velocity and a is acceleration [1]. Since the ball is dropped, v0 = 0, a = g = 9.8 m/s^2, and t = 3 s. Substituting gives v = 0 + (9.8 m/s^2)(3 s) = 29.4 m/s downward. Thus the numerical speed after 3 seconds is 29.4 m/s, corresponding to option 3. The calculation assumes uniform gravitational acceleration and neglects air resistance, consistent with standard free-fall kinematics [2].

Sources

1. [1] https://www2.tntech.edu/leap/murdock/books/v1chap2.pdf
2. [2] https://www1.grc.nasa.gov/beginners-guide-to-aeronautics/motion-of-free-falling-object/