Question

A boy standing at the top of A tower of20m height drops A stone. Assuming g= 10m/s² the velocity with which it hits the ground is

Answer 1

Answer:

$\large v=20 \ m/sec$

Explanation:

Given :

Initial velocity ( u ) = 0 m / sec

Height ( h ) = 20 m

Acceleration due to gravity ( g ) = 10 m / sec² .

We have to find final velocity ( v )

From third equation on motion when g is positive we have

$\large v^2=u^2+2gh$

Put the value here we get

$\large v^2=0^2+2\times10\times20\\\\\\\large v^2=400\\\\\\\large v=\sqrt{400}\\\\\\\large v=20 \ m/sec$

Thus we get final velocity .

Answer 2

» A boy standing at the top of A tower of 20 m height drops a stone.

Here..

• Initial velocity (u) = 0 m/s

(As no work done is done so, initial velocity is zero.)

• Height (h) = 20 m

• Acceleration (a) = 10 m/s²

____________ [ GIVEN ]

• We have to find the Velocity with which it hit the ground.

Means we have to find it's final velocity (v).

______________________________

We know that ..

v² - u² = 2as

(Third law of Motion)

Here ..

s = h and a = g

So;

→ v² - u² = 2gh

Put the known values in above formula

=> (v)² - (0)² = 2(10)(20)

=> v² = 400

=> v = √400

=> v = 20 m/s

_____________________________

The stone hits the ground with the velocity of 20 m/s.

__________ [ ANSWER ]

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