To tackle this problem, we need to analyze the motion of a body projected vertically upward with an initial velocity \( u \) while considering the effects of gravity and air resistance. The air resistance is constant and denoted as \( R \). We will derive the equations for the time of ascent and descent, as well as the relationship between the final velocity \( v \) and the initial velocity \( u \).
Understanding Forces in Motion
When the body is projected upward, two main forces act on it: the gravitational force \( mg \) (where \( m \) is the mass of the body and \( g \) is the acceleration due to gravity) and the air resistance \( R \). The net force acting on the body during ascent can be expressed as:
- Net Force = Upward Force - Downward Forces
- Net Force = \( -mg - R \)
This results in a negative acceleration (deceleration) as the body moves upward. According to Newton's second law, we can write:
\( ma = -mg - R \)
Deriving Time of Ascent
From the equation above, we can express the acceleration \( a \) as:
\( a = -g - \frac{R}{m} \)
Using the kinematic equation for velocity, we have:
\( v = u + at \)
At the peak of the ascent, the final velocity \( v \) becomes 0. Thus, we can set up the equation:
\( 0 = u - \left(g + \frac{R}{m}\right)t_a \)
Rearranging gives us:
\( t_a = \frac{u}{g + \frac{R}{m}} \)
Multiplying both sides by \( m \) leads to:
\( t_a = \frac{mu}{mg + R} \)
Time of Descent
Now, let's analyze the time of descent. When the body falls back to the ground, the forces acting on it change. The net force during descent is:
- Net Force = Weight - Air Resistance
- Net Force = \( mg - R \)
Applying Newton's second law again, we have:
\( ma = mg - R \)
Thus, the acceleration during descent is:
\( a = g - \frac{R}{m} \)
Using the same kinematic equation for velocity, we can express the final velocity \( v \) at the ground level:
\( v = 0 + \left(g - \frac{R}{m}\right)t_d \)
Rearranging gives us:
\( t_d = \frac{v}{g - \frac{R}{m}} \)
Multiplying both sides by \( m \) results in:
\( t_d = \frac{mV}{mg - R} \)
Relating Final and Initial Velocities
To find the relationship between the final velocity \( v \) and the initial velocity \( u \), we can use the conservation of energy or kinematic relationships. The velocity at the peak of ascent is 0, and we can relate the velocities using the forces acting on the body:
From the energy perspective, the work done against air resistance during ascent equals the change in kinetic energy:
\( \frac{1}{2}mu^2 = mgh + Rt_a \)
Substituting \( h = \frac{u^2}{2(g + \frac{R}{m})} \) into the equation and simplifying leads us to:
\( v/u = \sqrt{\frac{mg - R}{mg + R}} \)
Summary of Results
In summary, we have derived the following equations:
- Time of ascent: \( t_a = \frac{mu}{mg + R} \)
- Time of descent: \( t_d = \frac{mV}{mg - R} \)
- Velocity relationship: \( \frac{V}{u} = \sqrt{\frac{mg - R}{mg + R}} \)
These equations illustrate the impact of air resistance on the motion of the body both during ascent and descent, providing a comprehensive understanding of the dynamics involved.
