Physics, asked by drshanti23, 9 months ago

A stone is dropped into a well of 20 m deep.
Another stone is thrown downward with
velocity 'v' one second later. If both stones
reach the water surface in the well simultan-
eously, v is equal to
(g = 10 ms
1) 30 ms" 2) 15 ms 3) 20 ms! 4) 10 ms.
pls give clear answer pls with solution

Answers

Answered by toyashvipandey

Answer:

The time taken by the first stone to fall (in air to

the top of the well) 20 m below the drop point is

given by the relation,

s = ut + 12 g t?

In our case u= 0 m/s (as the stone is just

dropped), s= 20 m, taking g= 10 m/s2, we get from

the relation,

20 m = 0xt +12 x10 m/s2 x t? sSolving we get t= 2

It would takes the first stone 2s to fall 20m to the

top of the water in the well.

The second stone needs to cover these 20m in 1 s

( as it is thrown with a velocity v after the first

stone is dropped). Using the relation, and

substituting various values we get,

20 m = v m/s x 1s +12 x10 m/s2 x1 s2= y + 5 m

So,

v = 15 m/s.

Answered by Anonymous

Answer:

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A stone is dropped into a well of 20 m deep.Another stone is thrown downward withvelocity 'v' one second later. If both stonesreach the water surface in the well simultan-eously, v is equal to(g = 10 ms1) 30 ms" 2) 15 ms 3) 20 ms! 4) 10 ms. pls give clear answer pls with solution​

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A stone is dropped into a well of 20 m deep.
Another stone is thrown downward with
velocity 'v' one second later. If both stones
reach the water surface in the well simultan-
eously, v is equal to
(g = 10 ms
1) 30 ms" 2) 15 ms 3) 20 ms! 4) 10 ms.
pls give clear answer pls with solution