Question

A body freely failing from rest has a velocity V after it falls through a height h. The
distance it has to fall further for its velocity to become double is :
(a)3 h
(b) 6 h
(c) 8 h
Id 10h h​

Answer 1

ANSWER: (a)3h

Explanation:

if a body if falling from rest it's initial velocity u = 0m/s

and its given that it has a velocity V fter it falls a height be 'h' = Vm/s and a = 10m/s^2

therfore, by

V^2=u^2+2ah

V^2= 0+2 X 10 X h

V^2=20h

hence, here V becomes our initial velocity and V' become our final velocity as 2V =V'

(2V)^2 = V'^2

THEREFORE,

V'^2 = V^2 + 2 X a X H (where H is the height from V to V')

= (2V)^2 = V^2 + 2 X a X H

= 80h = 20h +2 X 10 X H

= H = 3h

hence our answer is 3h(a)

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