Question

a stone with weight w is thrown vertically upward into the air from ground level with initial speed vo if a force due to air drag acts on the stone throughout its flight . The maximum height attained by the stone is ​

Answer 1

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

Speed = ${V_0}$

Let the force due to air drag be "f".

$\huge{\bold{\underline{Explanation:-}}}$

As we know

$\bold{ \implies W = mg}$

Now,

$\bold{ \implies m = \frac{W}{g}}$

Now, Applying Newton's second law.

$\large{\bold{F = ma}}$

Here force will be

F = (W - f)

as Force due to air drag is acting opposite to the motion of the body.

$\large{ \implies(W - f) = ma}$

Here a is acceleration of the body moving up.

$\large{ \implies (W - f) = \frac{W}{g} \times a}$

$\large{\boxed{a = \frac{(W - f)g}{W}}}$

Applying Third kinematic equation.

$\large{\bold{v^2 - u^2 = 2as}}$

At highest point velocity will be zero.

$\large{ \implies 0 - (V_0^2) = 2aH_{max}}$

$\large{ \implies H_{max} = \frac{V_0^2}{2a}}$

Substituting the value of acceleration.

$\large{ \implies H_{max} = \frac{V_0^2}{2 \frac{(W - f)g}{W}}}$

$\huge{\boxed{\boxed{H_{max} = \frac{WV_0^2}{2(W - f)g}}}}$

This is the height attained by the stone.