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Zinaida [17]
4 years ago
14

2. A pebble is dropped down a well and hits the water 1.5 s later. Using the equations for motion with constant acceleration, de

termine the distance from the edge of the well to the water's surface.​
Physics
1 answer:
kow [346]4 years ago
3 0

Answer:

S = 11.025 m

Explanation:

Given,

The time taken by the pebble to hit the water surface is, t = 1.5 s

Acceleration due to gravity, g = 9.8 m/s²

Using the II equations of motion

                          S = ut + 1/2 gt²

Here u is the initial velocity of the pebble. Since it is free-fall, the initial velocity

                                u = 0

Therefore, the equation becomes

                            S = 1/2 gt²

Substituting the given values in the above equation

                              S = 0.5 x 9.8 x 1.5²

                                 = 11.025 m

Hence, the distance from the edge of the well to the water's surface is, S = 11.025 m

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The rate of flow of water is 71.28 kg/s

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A' = Cross sectional area of larger pipe

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Answer:

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= 6.91 x 10⁻³ T  .

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