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Grade: 11
        
A body freely falling from rest has a velocity v after it falls through distance h.The distance it has to fall down further for its velocity to becomes double is:
2 years ago

Answers : (3)

Arun
23820 Points
							
 
For first 2 m,
v^2 = 0^2 + 2g*2
v^2 = 4g
 
After that let it traverse d m futher more when its v becomes double .i.e. 2v
(2v)^2 = v^2 + 2gd
2gd = 3v^2
d = 3v^2/2g
put the value of v^2
d = 3*4g/2g = 12/2 = 6 meter.
 
Regards
Arun (askIITians forum expert)
2 years ago
Gowtham
40 Points
							
equation → v2-u2=2as
case 1--> free fall
u=0
v2=2gh
case2---from height h to h1
velocity changes from v to 2v
(2v)2-v2=2gh1
 
from the two equations we get relation between h and h1 as
h1=3*h
 
2 years ago
Khush
13 Points
							
Let height from which body is falling when v = b and u = 0 be h'1
Let height from which body is falling when v became double and u = v be h'2
2as=v²-u²(using equation 3 of motion
= 2×g×h'1=v²
= 2gh'1 = v².  @1
2gh'2=(2v)² - v². ( This is because we have to make v double to find answer)
=2gh'2=4v²-v²
=2gh'2=3v²
=h'2=3v²/2g (put valu of v² from @1)
=3×2gh'1/2g = h'2
=3h'1=h'2
Answer = 3h'1 in which v became double
 
 
 
 
 
 
 
 
 
10 months ago
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