a particle thrown vertically upwards with initial velocity u. find the speed it strikes back when there is air resiistance cu*2. where c is a positive constant and u is instantaneous speed.

To analyze the motion of a particle thrown vertically upwards with an initial velocity \( u \) while considering air resistance proportional to the square of its speed, we can break down the problem into several logical steps. The air resistance can be modeled as a force acting against the motion of the particle, which complicates the dynamics compared to a simple projectile motion scenario.

Understanding the Forces Involved

When the particle is thrown upwards, two main forces act on it:

• The gravitational force, which acts downwards and is equal to \( mg \), where \( m \) is the mass of the particle and \( g \) is the acceleration due to gravity.
• The air resistance, which acts upwards when the particle is moving upwards and is given by \( F_{\text{air}} = -c v^2 \), where \( c \) is a positive constant and \( v \) is the instantaneous speed of the particle.

Setting Up the Motion Equations

When the particle is moving upwards, the net force can be expressed as:

Net Force (upwards): \( F_{\text{net}} = -mg - c v^2 \)

According to Newton's second law, this can be written as:

Equation of Motion: \( m \frac{dv}{dt} = -mg - c v^2 \)

Finding the Velocity at the Peak

At the peak of its trajectory, the particle's velocity becomes zero. To find the time taken to reach this point, we can rearrange the equation:

Integrating the equation will give us the velocity as a function of time. However, for simplicity, we can focus on energy considerations or numerical methods to find the peak height and the subsequent descent.

Descent Phase

When the particle starts descending, the forces acting on it are still the same, but the direction of the air resistance changes. The net force during the descent can be expressed as:

Net Force (downwards): \( F_{\text{net}} = mg - c v^2 \)

Again, applying Newton's second law:

Equation of Motion: \( m \frac{dv}{dt} = mg - c v^2 \)

Energy Considerations

To find the speed at which the particle strikes back to the ground, we can use the principle of energy conservation, taking into account the work done against air resistance. The total mechanical energy at the initial point (when thrown) and just before it strikes the ground can be equated, considering the work done by air resistance.

Calculating the Final Speed

Let’s denote the speed just before impact as \( v_f \). The energy lost due to air resistance can be modeled as:

Work Done by Air Resistance: \( W = \int_{0}^{h} c v^2 \, dh \)

Where \( h \) is the height reached. The total mechanical energy at the peak will be the initial kinetic energy minus the work done against air resistance. This leads to a complex integral that can be solved numerically or through approximation methods.

Final Thoughts

In summary, the speed at which the particle strikes back to the ground can be determined by analyzing the forces acting on it during both ascent and descent, considering the work done against air resistance. The exact value of \( v_f \) will depend on the parameters \( u \) and \( c \), and solving the equations may require numerical methods or simulations for precise results. This approach illustrates the interplay between forces, energy, and motion in a real-world scenario where air resistance cannot be ignored.