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

To determine the maximum height attained by a stone thrown vertically upward with an initial speed \( v_0 \) while experiencing a constant air drag force \( f \), we can analyze the forces acting on the stone and apply the principles of physics, specifically Newton's second law and energy conservation.

Understanding the Forces at Play

When the stone is thrown upward, two main forces act on it:

• The gravitational force acting downward, which is equal to \( mg \) (where \( m \) is the mass of the stone and \( g \) is the acceleration due to gravity).
• The air drag force \( f \), which also acts downward, opposing the stone's motion.

Net Force Calculation

The net force \( F_{\text{net}} \) acting on the stone can be expressed as:

F_net = -mg - f

Here, both forces are negative because they act in the opposite direction to the initial velocity of the stone. According to Newton's second law, we can write:

F_net = ma

Where \( a \) is the acceleration of the stone. Setting these equal gives us:

ma = -mg - f

From this, we can find the acceleration:

a = -g - \frac{f}{m}

Using Kinematics to Find Maximum Height

To find the maximum height, we can use the kinematic equation that relates initial velocity, final velocity, acceleration, and displacement:

v^2 = v_0^2 + 2a h

At the maximum height, the final velocity \( v \) is 0. Plugging in the values, we get:

0 = v_0^2 + 2(-g - \frac{f}{m})h

Rearranging this equation to solve for height \( h \) gives:

h = \frac{v_0^2}{2(g + \frac{f}{m})}

Final Expression for Maximum Height

This formula shows that the maximum height \( h \) attained by the stone depends on its initial speed \( v_0 \), the gravitational acceleration \( g \), and the air drag force \( f \) relative to the mass \( m \) of the stone. The greater the initial speed, the higher the stone will go, but the presence of air drag reduces this height.

Example Calculation

Let’s say we have a stone with a mass of 2 kg, an initial speed of 20 m/s, and an air drag force of 5 N. The acceleration due to gravity is approximately 9.81 m/s². Plugging these values into our height formula:

h = \frac{20^2}{2(9.81 + \frac{5}{2})}

Calculating the denominator:

h = \frac{400}{2(9.81 + 2.5)} = \frac{400}{2 \times 12.31} = \frac{400}{24.62} \approx 16.24 \text{ meters}

This example illustrates how to apply the concepts of forces and kinematics to find the maximum height of a stone thrown upward while considering air drag. Understanding these principles can help you analyze similar problems in physics effectively.