Physics, asked by stuf2013109942, 5 months ago

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 MystícPhoeníx
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 arandombackslash
0

Answer:

19.6 m/s

Explanation:

Use the equations of motion for solving this problem, firstly,

pick the first equation of motion

v=u+at

Inserting values, into it, we get,

v = u+9.8\cdot2

note that, u is the initial velocity, which is 0, as it starts falling down from the rest stage, since it was thrown, hence

v=0 + 19.6

v=19.6

Hence the velocity is 19.6 m/s

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