A body freely falling from the rest has a velocity $‘v’$ after it falls through a height $‘h’$. The distance it has to fall down for its velocity to become double, is
A body freely falling from the rest has a velocity $‘v’$ after it falls through a height $‘h’$. The distance it has to fall down for its velocity to become double, is
$2h$
$4h$
$6h$
$8h$
A body freely falling from the rest has a velocity $‘v’$ after it falls through a height $‘h’$. The distance it has to fall down for its velocity to become double, is
Let at point $A$ initial velocity of body is equal to zero
for path $AB$ : ${v^2} = 0 + 2gh$…(i)
for path $AC $ : ${(2v)^2} = 0 + 2gx$
$4{v^2} = 2gx$…(ii)
Solving (i) and (ii)
$x = 4h$
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