Physics, asked by Jeneliya27, 1 year ago

A body falling from rest has velocity v after it falls for distance h calculate the distance it has to fall down further for its velocity to become quadruple times h????? plz guys solve this fast. l surely mark as brainaist....to....

Answers

Answered by Aviral101
2
initial u = 0
final v = v
Distance= h
v^2 = 2gh ----1

Now, v=4v, let h be s
(4v)^2 = 2gs
16v^2 = 2gs. ----2

by 1, 2
v^2 is 16 timed.
2gh:2gs = 1:16

therefore the more it has to travel = 16h - h
= 15h

Aviral101: pls mark as brainliest
Answered by t6anush
0

Answer:

Explanation:

Here, initial velocity of the body v  

1

​  

=v

Initial height h = hc

Final velocity of the body v  

2

​  

=2v

Now, from the equation of motion  

v  

2

=u  

2

+2gh

v  

2

∝h

So, [  

v  

2

​  

 

v  

1

​  

 

​  

 

​  

]  

2

=[  

h  

2

​  

 

h  

1

​  

 

​  

 

​  

]

or, [  

2v

v

​  

 

​  

]  

2

=  

h  

2

​  

 

h

​  

 

or,  

h  

2

​  

 

h

​  

=  

4

1

​  

 

or h  

2

​  

=4h

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