37 39 and 40 question
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Using equation
s = ut + (at^2)/2
u = 0
so
s = (at^2)/2
So, time is directly proportional to √s
s is height in the given case.
Q37 ans is 'a' part.
Q39 ans is 'a' part.
In Q40,
x(2) = 3*2 - 4*2*2 + 2*2*2= -2
x(4) = 3*4 - 4*4*4 + 4*4*4 = 12
∆t = 4 - 2 = 2
Velocity = Displacement/∆t
= {x(4) - x(2)}/∆t
= (12 - (-2))/2
= 14/2
= 7m/s
So in Q41 option 'a' is correct.