Question

A stone with weight w is thrown vertically upward into air from ground with velocity U.
If a constant force f due to air drag acts on the stone, then find maximum height attained.

Answer 1

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● A stone with weight w is thrown vertically upward into air from ground with velocity U.

A stone with weight w is thrown vertically upward into air from ground with velocity U.If a constant force f due to air drag acts on the stone, then find maximum height attained?

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●We know,

$w = work \\ f = force$

$w + f = ma - - - - (1)$

$m = \frac{w}{g}$

$putting \: the \: value \: of \: m \: in \: 1st \: equation$

$w + f = \frac{w}{g} \times a$

$a = \frac{w + f}{w} \times g$

$we \: know$

$h(max) = \frac{ {v}^{2} }{2a}$

$\red{\boxed </p><p>{\frac{ {v}^{2} }{2g(1 + \frac{f}{w}) }}}$

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Answer 2

||✪✪ QUESTION ✪✪||

A stone with weight w is thrown vertically upward into air from ground with velocity U.

If a constant force f due to air drag acts on the stone, then find maximum height attained.

|| ✰✰ ANSWER ✰✰ ||

Let us assume that, the stone is reaching the maximum height upto h metre.

Now, using conservation of Energy we get ,,,

→ initial Energy = (1/2) mu²

And, at Highest Point Velocity is ZERO.. So, Final Kinetic Energy will also Becomes Zero..

Now,

kinetic Energy is using against work done and air Friction.

→ work done is = mgh

→ Friction = constant force * Maximum height = f * h

So, we can say that :-

==>> (1/2) mu² = mgh + f * h

Now, using mg = w (weight) in RHS, we get,

==>> (1/2) mu² = h (w + f)

Now, in LHS, Multiply and divide by g, we get,

==>> (1/2) (mg)u²/g = h (w + f)

Again , using mg = w now, in LHS,

==>> (1/2) * w * u²/g = h (w + f)

==>> h = [ w * u² ] / [ 2g (w + f) ] (Ans).