For a body thrown vertically upwards, if the air resistance is taken into consideration, then the time of rise is

Options

A. equal to time of fall
B. less than time of fall
C. greater than time of fall
D. (3/2) times the time of fall

Answer

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

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Body**:
When a body is thrown upwards, two main forces act on it:
- The gravitational force (weight) acting downwards, which is \( mg \).
- The air resistance (drag) acting upwards, which we can denote as \( F \).

2. **Establish the Initial Conditions**:
Let the initial velocity of the body be \( u \) and the final velocity at the highest point be \( 0 \). The height reached by the body is \( h \).

3. **Determine the Effective Acceleration During Ascent**:
When the body is moving upwards, the net force acting on it can be expressed as:
\[ F_{\text{net}} = mg + F \]
Therefore, the effective acceleration \( a_1 \) during the upward motion is:
\[ a_1 = g + \frac{F}{m} \]
This indicates that the body experiences a greater effective acceleration due to both gravity and air resistance.

4. **Determine the Effective Acceleration During Descent**:
When the body is falling back down, the forces acting on it are:
- The gravitational force \( mg \) acting downwards.
- The air resistance \( F \) acting upwards.

The net force during the downward motion is:
\[ F_{\text{net}} = mg - F \]
Thus, the effective acceleration \( a_2 \) during the downward motion is:
\[ a_2 = g - \frac{F}{m} \]
This indicates that the body experiences a lesser effective acceleration when descending due to the opposing air resistance.

5. **Use Kinematic Equations**:
The distance \( h \) traveled during both the ascent and descent can be described by the kinematic equation:
\[ h = ut + \frac{1}{2} a t^2 \]
For the upward motion, we can express it as:
\[ h = ut_{\text{up}} - \frac{1}{2} (g + \frac{F}{m}) t_{\text{up}}^2 \]
For the downward motion, it can be expressed as:
\[ h = \frac{1}{2} (g - \frac{F}{m}) t_{\text{down}}^2 \]

6. **Analyze Time of Ascent vs. Time of Descent**:
Since the distance \( h \) is the same for both ascent and descent, we can conclude that:
- The time of rise \( t_{\text{up}} \) will be less than the time of fall \( t_{\text{down}} \) because the effective acceleration during ascent is greater than during descent.

7. **Conclusion**:
Therefore, the time of rise is less than the time of fall when air resistance is taken into account.

### Final Answer:
The time of rise is less than the time of fall.