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A ball is dropped from the top of a high building with a constant accleration 9.8m/s2. What will be its velocity after 3 seconds?
- a)9.8 m/s
- b)19.6 m/s
- c)29.4 m/s
- d)39.2 m/s
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
A ball is dropped from the top of a high building with a constant accl...
The velocity after time t is given by
v= u + gt
Here, u = 0, t = 2s, g = 9.8 m/s�
v = 0 + (9.8 x 2) = 19.6 m/s
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Community Answer
A ball is dropped from the top of a high building with a constant accl...
Given information:
- Acceleration of the ball: 9.8 m/s^2
- Time: 3 seconds
To find the velocity of the ball after 3 seconds, we can use the equation of motion:
v = u + at
Where:
v is the final velocity
u is the initial velocity (in this case, it is 0 m/s since the ball is dropped)
a is the acceleration
t is the time
Calculating the velocity:
v = 0 + (9.8)(3)
v = 0 + 29.4
v = 29.4 m/s
Therefore, the velocity of the ball after 3 seconds is 29.4 m/s.
Final Answer:
The correct option is c) 29.4 m/s.
- Acceleration of the ball: 9.8 m/s^2
- Time: 3 seconds
To find the velocity of the ball after 3 seconds, we can use the equation of motion:
v = u + at
Where:
v is the final velocity
u is the initial velocity (in this case, it is 0 m/s since the ball is dropped)
a is the acceleration
t is the time
Calculating the velocity:
v = 0 + (9.8)(3)
v = 0 + 29.4
v = 29.4 m/s
Therefore, the velocity of the ball after 3 seconds is 29.4 m/s.
Final Answer:
The correct option is c) 29.4 m/s.
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A ball is dropped from the top of a high building with a constant accleration 9.8m/s2. What will be its velocity after 3 seconds?a)9.8 m/sb)19.6 m/sc)29.4 m/sd)39.2 m/sCorrect answer is option 'C'. Can you explain this answer?
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