Question

A freely falling body has velocity v after covering a distance of h, what will be the velocity after covering a total distance of 2h from the point where its velocity was u=0? Photographic explanation please

Answer 1

Answer:

Initial speed of the body u=0

Acceleration of the freely falling body  a=g

Velocity of body after covering h distance,  v  

1

​  

=V

Using  v  

1

​  

−u  

2

=2gh

∴  V  

2

−0=2gh

We get  V  

2

=2gh                  ....(1)

Let the total distance covered by body to attain v  

2

​  

=2V be H.

Using  v  

2

​  

−u  

2

=2gH

Or   4V  

2

−0=2gH

Or  4V  

2

=2gH

Or 4(2gh)=2gH

Thus we get  H=4h

Now required distance the body needed to fall Δh=H−h

∴  Δh=4h−h=3h

Answer 2

Explanation:

Initial speed of the body u=0

Acceleration of the freely falling body a=g

Velocity of body after covering h distance, v1 =V

Using v1 −u^2 =2gh

∴ V^2−0=2gh

We get V^2 =2gh ....(1)

Let the total distance covered by body to attain v

2

=2V be H.

Using v^2-u^2 = 2gH

Or 4V^2−0=2gH

Or 4V^2 =2gH

Or 4(2gh)=2gH

Thus we get H=4h

Now required distance the body needed to fall Δh=H−h

∴ Δh=4h−h=3h

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