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....
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Answered by
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
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
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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