A body is thrown vertically upwards. If air resistance is to be taken into account, then the time during which the body rises is.

Options

A. Equal to the time of fall
B. Less than the time of fall
C. Greater than the time of fall
D. Twice the time of fall

Explanation

To solve the problem of a body thrown vertically upwards considering air resistance, we can follow these steps:

### Step 1: Understand the Forces Acting on the Body
When a body is thrown upwards, two forces act on it:
1. The gravitational force (downward) which is equal to \( mg \).
2. The air resistance (also downward) which we will denote as \( F_a \) and is assumed to be constant for this problem.

### Step 2: Write the Equation of Motion
The net force acting on the body when it is moving upwards is:
\[ F_{net} = -mg - F_a \]
This gives us the acceleration \( a \) of the body as:
\[ a = -g - a \]
where \( a \) is the magnitude of air resistance per unit mass.

### Step 3: Use Kinematic Equations
The initial velocity of the body is \( u \), and we want to find the time \( t_1 \) taken to reach the maximum height where the final velocity \( v = 0 \). Using the equation:
\[ v = u + at \]
At maximum height, \( v = 0 \):
\[ 0 = u - (g + a)t_1 \]
Rearranging gives:
\[ t_1 = \frac{u}{g + a} \]

### Step 4: Analyze the Time of Fall
When the body falls back down, the forces acting on it are:
1. Gravitational force (downward) \( mg \).
2. Air resistance (upward) \( F_a \).

The net acceleration during the fall is:
\[ a_{fall} = g - a \]
The initial velocity for the fall is \( 0 \) (at the maximum height). We can use the same kinematic equation:
\[ h = ut + \frac{1}{2} a_{fall} t_2^2 \]
Since \( u = 0 \) for the fall, we have:
\[ h = \frac{1}{2} (g - a) t_2^2 \]

### Step 5: Calculate the Height
From the upward motion, we can find the height \( h \):
\[ h = \frac{u^2}{2(g + a)} \]

### Step 6: Set the Heights Equal
Since the height reached during the rise is equal to the height fallen, we can set the two height equations equal:
\[ \frac{u^2}{2(g + a)} = \frac{1}{2} (g - a) t_2^2 \]
Cancelling \( \frac{1}{2} \) gives:
\[ \frac{u^2}{g + a} = (g - a) t_2^2 \]

### Step 7: Solve for \( t_2 \)
Rearranging gives:
\[ t_2^2 = \frac{u^2}{(g + a)(g - a)} \]
Taking the square root:
\[ t_2 = \frac{u}{\sqrt{(g + a)(g - a)}} \]

### Step 8: Compare \( t_1 \) and \( t_2 \)
Now we have:
- Time of rise: \( t_1 = \frac{u}{g + a} \)
- Time of fall: \( t_2 = \frac{u}{\sqrt{(g + a)(g - a)}} \)

To compare \( t_1 \) and \( t_2 \), we can analyze the ratios:
\[ \frac{t_2}{t_1} = \frac{(g + a)}{\sqrt{(g + a)(g - a)}} \]
Since \( g + a > g - a \), it follows that \( t_2 > t_1 \).

### Conclusion
Thus, the time during which the body rises \( t_1 \) is less than the time during which it falls \( t_2 \).