Question

A stone with weight w is thrown vertically upward into the air from ground level with initial speed v 0 ​ . If a constant force f due to air drag acts on the stone throughout its flight. The maximum height attained by the stone is

Answer 1

v_0^2 / (2g(1 + f/w))

Answer 2

The problem asks for the maximum height attained by a stone thrown vertically upward with initial speed $v_0$, considering its weight $w$ and a constant air drag force $f$.

1. Identify forces during upward motion:

When the stone is moving vertically upward, its velocity is directed upward. The forces acting on the stone are:

• Weight ($w$): Acts vertically downward.
• Air drag force ($f$): Acts vertically downward, opposing the upward motion.

2. Calculate the net force and acceleration during upward motion:

The total downward force (net force) acting on the stone during its upward motion is:

$F_{net} = w + f$

According to Newton's second law, $F_{net} = ma$, where $m$ is the mass of the stone and $a$ is its acceleration.

So, $ma = w + f$.

The acceleration of the stone during its upward journey is $a = \frac{w+f}{m}$. This acceleration is directed downward.

We know that weight $w = mg$, where $g$ is the acceleration due to gravity. From this, the mass of the stone is $m = \frac{w}{g}$.

Substitute the mass $m$ into the acceleration equation:

$a = \frac{w+f}{w/g} = g \left( \frac{w+f}{w} \right) = g \left( 1 + \frac{f}{w} \right)$

This is the effective downward acceleration experienced by the stone during its upward flight.

3. Use kinematic equation for maximum height:

To find the maximum height ($H_{max}$), we use the kinematic equation for uniformly accelerated motion:

$v^2 = u^2 + 2aS$

Here:

• Initial velocity ($u$) = $v_0$ (upward)
• Final velocity ($v$) = $0$ (at maximum height, the stone momentarily stops)
• Acceleration ($a$) = $-g \left( 1 + \frac{f}{w} \right)$ (taking the upward direction as positive, the acceleration is downward, hence negative)
• Displacement ($S$) = $H_{max}$ (the maximum height attained)

Substitute these values into the kinematic equation:

$0^2 = v_0^2 + 2 \left( -g \left( 1 + \frac{f}{w} \right) \right) H_{max}$

$0 = v_0^2 - 2g \left( 1 + \frac{f}{w} \right) H_{max}$

Rearrange the equation to solve for $H_{max}$:

$2g \left( 1 + \frac{f}{w} \right) H_{max} = v_0^2$

$H_{max} = \frac{v_0^2}{2g \left( 1 + \frac{f}{w} \right)}$

The maximum height attained by the stone is $\frac{v_0^2}{2g \left( 1 + \frac{f}{w} \right)}$.