Question

A stone dropped from top of a well reaches the surface of water in 2 seconds,find the velocity of stone while it touches the surface of water and what is the depth of the water surface from top of well(g=10m/S2) (




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Answer 1

Answer:

v=u+gt

v=10×2

=20m/s

v^2=u^2+2gh

20^2=2×10×h

h=400/20

=20m

Answer 2

Answer:

The velocity of the stone while it touches the water surface is 20 m/s

While the depth of water surface is 20m.

Explanation:

Using equation

$h = 5 {t}^{2}$

putting the value of time = 2 sec

=> the depth of water from top surface = 5 × 4 = 20 meter

Using another equation

$v = \sqrt{20h}$

putting h= 20 meter

=> velocity of stone when it touches water = square root of 400 = 20 m/s

YOU WOULD WONDER FROM WHERE THESE TWO EQUATION CAME

So the answer is simple that I have derived these from the three simple equations of motion, HOW ?

see yourself in the pic not only these two but Four more also.

(I'm showing you these because I love helping friends, thanks for your support)