Question

a stone dropped from top of a well reaches the surface of water in 2s 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/s^2) (using v=u+at,s=ut+1/2at^2​

Answer 1

Answer:

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 - 117… ... 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)​.

Explanation:

Answer 2

GiveN :

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

Therefore,

• Initial velocity, u = 0 m/s [ball was initially at rest]

• Time taken, t = 2 s

• Final velocity, v= v

• Acceleration due to gravity, g = 10 m/s²

As per the question,

We've to solve this using the kinematic equations.

By using the first kinematic equation :

↬ v = u + gt

↬ v = 0 + 10(2)

↬ v = 20 m/s

Aa we've got the final velocity, we can easily find the height of the stone when it touches surface of water.

By using third kinematic equation :

$\longrightarrow \sf h = ut + \dfrac{1}{2} gt^2$

Substituting the values,

$\longrightarrow \sf h = 0(2) + \dfrac{1}{2} \times 10 \times 2 \times 2$

$\longrightarrow \sf h = 0 + 10 \times 2$

$\longrightarrow \large{ \boxed {\sf h = 20 \ m }}$

$\therefore$ Velocity of the stone is 20 m/s & height of the stone when it touches surface of water is 20 m.