Physics, asked by nightfury27, 6 months ago

A stone with weight ‘W’ is thrown vertically upward into the air from ground level with initial speed ‘u’. If a constant force ‘F’ due to air drag acts on the stone throughout its flight. The maximum height attained by the stone is

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Answered by Anonymous

Answer:

According to the problem the weight of the stone is w

therefore the mass will be m = w/g

As we know the air drops any particle in the opposite direction of its velocity therefore for this case,

w/g x a = w - f

here a is the acceleration of the stone with its velocity towards up

=> a = (w-f) x g/w

Let the maximum height attained is h

Now from equation of motion we can say that,

0 = v0^2 - 2ah

h = v0^2/2a

= v0^2/ 2 (w-f) x g/w

= wv0^2/2(w−f) x g

Explanation:

Answered by niishaa

According to the problem the weight of the stone is w

therefore the mass will be m = w/g

As we know the air drops any particle in the opposite direction of its velocity therefore for this case,

w/g x a = w - f

here a is the acceleration of the stone with its velocity towards up

=> a = (w-f) x g/w

Let the maximum height attained is h

Now from equation of motion we can say that,

0 = v0² - 2ah

h = v0²/2a

= v0²/ 2 (w-f) x g/w

= wv0²/2(w−f) x g

Explanation:

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A stone with weight ‘W’ is thrown vertically upward into the air from ground level with initial speed ‘u’. If a constant force ‘F’ due to air drag acts on the stone throughout its flight. The maximum height attained by the stone is ​

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A stone with weight ‘W’ is thrown vertically upward into the air from ground level with initial speed ‘u’. If a constant force ‘F’ due to air drag acts on the stone throughout its flight. The maximum height attained by the stone is