A ball is dropped over the edge of a building. How fast is the ball moving 2.0 seconds after being
dropped?
Answers
Answered by
15
Given:-
- Initial velocity, u = 0m/s
- Acceleration due to gravity = 9.8m/s²
- Time taken ,t = 2 s
To Find:-
- Final velocity ,v
Solution:-
By using 1st equation of motion
• v = u +at
Substitute the value we get
→ v = 0 + 9.8×2
→ v = 19.6 m/s
Therefore the Final velocity of the ball is 19.6 m/s.
Additional Information!!
• The rate of change of velocity at per unit time is called acceleration .
• SI unit of acceleration is m/s²
• It is vector Quantity ( both magnitude and direction)
Answered by
0
Answer:
19.6 m/s
Explanation:
Use the equations of motion for solving this problem, firstly,
pick the first equation of motion
Inserting values, into it, we get,
note that, u is the initial velocity, which is 0, as it starts falling down from the rest stage, since it was thrown, hence
Hence the velocity is 19.6 m/s
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