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Answer 1

total acceleration on upward moving particle

=(weight of stone +force of air drag). ×1/mass

=g+f/m. (taking downward as positive)

now

max.height reached by body =V^2/2a

=v^2/2(g+f/m)

taking g as common we get

h=V^2/2g(1+f/mg)=V^2/2g(1+f/w)

so option 1 is correct

hope it helps you dude

Answer 2

SOLUTION

W = mg........(1)

where m is the mass of the body...

As you get the solution of this question so I provide u a another way to solve such type of questions...

We know that in this case the net force which is responsible for retardation of the stone is f + w

because this force is downward but the motion of the block is upwards.

-ma = f + w

where this negative sign indicate the direction of the retardation

-a = (f + w)/m

We can write a = vdv/dx

-vdv = (f + w)dx/m

integrate both side we get....

-[v²/2] = (f + w)[X]/m

limits of velocity goes from v at initial moment and to 0 at max. height

limits of X goes form 0 at initial moment to height h (max.)

-(0² - v²)/2 = (f + w)(h - 0)/m

v²/2(f + w) = h/m

v²m/2(f + w) = h

multiply and divide by g

v²mg/2g(f + w) = h. (mg = w)

v²w/2g(f + w) = h

v²/2g(f/w + 1) = h

Option à

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