A stone is projected with speed `20 m//s` at angle `37^(@)` with the horizontal and it hits the ground with speed `12m//s` due to air resistance. Assume the effect of air resistance to reduce only horizontal component of velocity. Then the time of flight will be-

Options

A. greater than `2.4 sec`
B. less than `2.4 sec`
C. `2.4 sec`
D. depends on other data

Explanation

To solve the problem, we will break it down into steps:

### Step 1: Identify the Given Information
- Initial speed of the stone, \( u = 20 \, \text{m/s} \)
- Angle of projection, \( \theta = 37^\circ \)
- Final speed when it hits the ground, \( v = 12 \, \text{m/s} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

### Step 2: Calculate the Components of the Initial Velocity
The initial velocity can be resolved into horizontal and vertical components:
- Horizontal component, \( u_x = u \cos \theta = 20 \cos 37^\circ \)
- Vertical component, \( u_y = u \sin \theta = 20 \sin 37^\circ \)

Using the trigonometric values:
- \( \cos 37^\circ = \frac{4}{5} \)
- \( \sin 37^\circ = \frac{3}{5} \)

Calculating the components:
- \( u_x = 20 \times \frac{4}{5} = 16 \, \text{m/s} \)
- \( u_y = 20 \times \frac{3}{5} = 12 \, \text{m/s} \)

### Step 3: Determine the Time of Flight
The time of flight for a projectile is determined by the vertical motion, which is independent of the horizontal motion affected by air resistance. The formula for the time of flight \( t \) is given by:
\[ t = \frac{2 u_y}{g} \]

Substituting the values:
\[ t = \frac{2 \times 12}{10} = \frac{24}{10} = 2.4 \, \text{s} \]

### Final Answer
The time of flight of the stone is \( 2.4 \, \text{s} \).

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