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A lift with open top is moving up with acceleration f. A body is thrown up with speed u relative to lift in time t. then speed u is [g= acceleration due to

to gravity]
1) ft+gt .... 2) √[(ft)²+(gt)²] .... 3) ft+gt/2 .... (4) f²t/g


Answer is A.....................

initial acceleration = f
at time t, velocity of lift = ft
at time t, velocity due to gravity = gt
then ball is thrown upward with velocity u, relative to lift
from ground, u = ft+gt
1 1 1
Please post the solution..or give a hint to solve
@Jahnavi97, Please post the solution..or give a hint to solve
[email protected]
It's actually wrong..no offense..... I solved it now...vel of body relative to lift is given [u]... acc due to gravity is [-g]...relative acc = [-g-f], since both are in same direction...Time of flight [t] = 2u/[g plus f], took the magnitude of relative acc because time can't be negative.... =>u = t*[g plus f]/2 ==> u = [gt plus ft]/2
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