Q: 114 (IAS/1998)
question_subject:
Science
question_exam:
IAS
stats:
0,11,21,18,2,11,1
keywords:
{'velocity': [0, 2, 2, 6], 'constant acceleration': [0, 0, 0, 1], 'high building': [0, 1, 0, 0], 'ball': [1, 3, 13, 12], 's2': [0, 1, 2, 6], 'seconds': [3, 3, 8, 6]}
The velocity of the ball after 3 seconds can be determined using the equation:
v = u + at
Where:
v = final velocity
u = initial velocity (which is 0 in this case as the ball is dropped)
a = acceleration (constant and equal to 9.8 m/s^2 in this case)
t = time
Plugging in the values, we have:
v = 0 + (9.8 m/s^2) * (3 s)
v = 0 + 29.4 m/s
v = 29.4 m/s
Therefore, the velocity of the ball after 3 seconds will be 29.4 m/s.