A particle is projected with speed `10 m//s` at angle `60^(@)` with the horizontal. Then the time after which its speed becomes half of initial.

Options

A. `1/2` sec
B. `1` sec
C. `sqrt(3//2) sec`
D. `sqrt(3)//2 sec`

Explanation

To solve the problem of finding the time after which the speed of a particle projected at an angle becomes half of its initial speed, we can follow these steps:

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

### Step 2: Calculate the horizontal and vertical components of the initial velocity
- The horizontal component of the velocity \( u_x \):
\[ u_x = u \cos \theta = 10 \cos 60^\circ = 10 \times \frac{1}{2} = 5 \, \text{m/s} \]

- The vertical component of the velocity \( u_y \):
\[ u_y = u \sin \theta = 10 \sin 60^\circ = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3} \, \text{m/s} \]

### Step 3: Determine the time when the speed becomes half of the initial speed
- The initial speed is \( 10 \, \text{m/s} \), so half of the initial speed is \( 5 \, \text{m/s} \).
- The speed of the particle at any time \( t \) can be calculated using the Pythagorean theorem:
\[ v = \sqrt{u_x^2 + v_y^2} \]
where \( v_y \) is the vertical component of the velocity at time \( t \).

### Step 4: Calculate the vertical component of the velocity at time \( t \)
- The vertical component of the velocity \( v_y \) at time \( t \) is given by:
\[ v_y = u_y - g t = 5\sqrt{3} - 9.8 t \]

### Step 5: Set up the equation for the speed
- We want the speed to be \( 5 \, \text{m/s} \):
\[ 5 = \sqrt{u_x^2 + v_y^2} \]
Substituting \( u_x \) and \( v_y \):
\[ 5 = \sqrt{(5)^2 + (5\sqrt{3} - 9.8t)^2} \]

### Step 6: Square both sides to eliminate the square root
\[ 25 = 25 + (5\sqrt{3} - 9.8t)^2 \]
\[ 0 = (5\sqrt{3} - 9.8t)^2 \]

### Step 7: Solve for \( t \)
- Since the square of a number is zero, we can set the expression inside the square to zero:
\[ 5\sqrt{3} - 9.8t = 0 \]
\[ 9.8t = 5\sqrt{3} \]
\[ t = \frac{5\sqrt{3}}{9.8} \]

### Step 8: Calculate the numerical value of \( t \)
Using \( \sqrt{3} \approx 1.732 \):
\[ t \approx \frac{5 \times 1.732}{9.8} \approx \frac{8.66}{9.8} \approx 0.88 \, \text{s} \]

Thus, the time after which the speed becomes half of the initial speed is approximately \( 0.88 \, \text{s} \).

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