Question

4) A stone dropped from the top of a well reaches the surface of water in 2
econds, find the velocity of stone while it touches the surface of water,​

Answer 1

Answer:

Here Initial velocity (u) = 58.8 m/s

Final velocity (v) = 0 (as it comes to rest on touching the ground)

Time (t) = 6 seconds

We know v = u - at

0 = 58.8 - a × 6

⇒6a=58.8

⇒a=

6

58.8

∴a=9.8m/s

2

∴ The value of acceleration due to gravity = 9.8 m/s

2

Answer 2

Answer:

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

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