A particle is projected at an angle `60^@` with horizontal with a speed `v = 20 m//s`. Taking `g = 10m//s^2`. Find the time after which the speed of the particle remains half of its initial speed.

Explanation

To solve the problem of finding the time after which the speed of the particle remains half of its initial speed, we can follow these steps:

### Step 1: Identify the initial conditions
- The initial speed \( u = 20 \, \text{m/s} \)
- The angle of projection \( \theta = 60^\circ \)
- The acceleration due to gravity \( g = 10 \, \text{m/s}^2 \)

### Step 2: Determine the speed when it is half of the initial speed
The speed at which we want to find the time is half of the initial speed:
\[ v = \frac{u}{2} = \frac{20}{2} = 10 \, \text{m/s} \]

### Step 3: Use the formula for the speed of a projectile
The speed \( v \) of a projectile at any time \( t \) can be expressed as:
\[ v = \sqrt{u^2 + (gt)^2 - 2u g t \sin \theta} \]
Substituting \( u = 20 \, \text{m/s} \), \( g = 10 \, \text{m/s}^2 \), and \( \theta = 60^\circ \) (where \( \sin 60^\circ = \frac{\sqrt{3}}{2} \)):
\[ 10 = \sqrt{20^2 + (10t)^2 - 2 \cdot 20 \cdot 10 \cdot t \cdot \frac{\sqrt{3}}{2}} \]

### Step 4: Square both sides to eliminate the square root
Squaring both sides gives:
\[ 100 = 400 + 100t^2 - 200t \cdot \sqrt{3} \]

### Step 5: Rearrange the equation
Rearranging the equation:
\[ 100t^2 - 200\sqrt{3}t + 400 - 100 = 0 \]
\[ 100t^2 - 200\sqrt{3}t + 300 = 0 \]

### Step 6: Simplify the equation
Dividing the entire equation by 100:
\[ t^2 - 2\sqrt{3}t + 3 = 0 \]

### Step 7: Solve the quadratic equation
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1 \), \( b = -2\sqrt{3} \), and \( c = 3 \):
\[ t = \frac{2\sqrt{3} \pm \sqrt{(2\sqrt{3})^2 - 4 \cdot 1 \cdot 3}}{2 \cdot 1} \]
\[ t = \frac{2\sqrt{3} \pm \sqrt{12 - 12}}{2} \]
\[ t = \frac{2\sqrt{3}}{2} = \sqrt{3} \]

### Step 8: Final answer
Thus, the time after which the speed of the particle remains half of its initial speed is:
\[ t \approx 1.732 \, \text{seconds} \]