Question

A stone dropped from top of a well reaches the surface of water 2seconds, 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)(using V=U+at,S=Ut+1/2 at2)​

Answer 1

Answer:

velocity of the stone

= 20 m/s

Height

= 20 meter

Step by step explanations :

Given that,

A stone dropped from top of a well reaches the surface of water 2 seconds

Here

Initial velocity of the stone = 0 m/s

[ball was initially at rest]

time taken to reach the water = 2 s

let the velocity of the stone when it touches the water surface be v

now we have,

initial velocity(u) = 0 m/s

final velocity(v) = v

time taken(t) = 2 s

gravitational acceleration(g) = 10 m/s²

by the gravitational equation of motion,

v = u + gt

putting the values,

v = 0 + 10(2)

v = 20 m/s

so,

final velocity of the stone = 20 m/s

Now,

let the height of the water from the initial position of the stone be h

h = ut + ½ gt²

h = 0(2) + ½ × 10 × 2 × 2

h = 20 m

so,

Height = 20 meter

_________________

velocity of the stone

= 20 m/s

Height

= 20 meter

Answer 2

Answer:

velocity of the stone = 20 m/s

Height = 20 meter

Step by step explanations :

Given,

Initial velocity of stone(u) = 0 m/s

time taken to reach water surface = 2 s

let the velocity of the stone by touching the water surface be v

now, m/s²

by the gravitational equation of motion,

v = u + gt

g = gravitational acceleration = 10 m/s²

putting values,

v = 0 + 10(2)

v = 20 m/s

so,

final velocity of stone = 20 m/s

Now,

h = ut + ½ gt²

where,

h = height of stone from water surface

so,

h = 0(2) + ½ × 10 × 2 × 2

h = 20 m

so,

Height = 20 meter