A particle is projected with speed `20m s^(-1)` at an angle `30^@` With horizontal. After how much time the angle between velocity and acceleration will be `90^@`

Options

A. 1 s
B. 2 s
C. 1.5 s
D. never

Answer

To solve the problem, we need to find the time after which the angle between the velocity and acceleration of a projectile becomes 90 degrees. This occurs at the maximum height of the projectile's trajectory.

### Step-by-Step Solution:

1. **Identify the initial conditions:**
- Initial speed \( u = 20 \, \text{m/s} \)
- Angle of projection \( \theta = 30^\circ \)

2. **Calculate the vertical component of the initial velocity:**
\[ u_y = u \sin \theta = 20 \sin 30^\circ = 20 \times \frac{1}{2} = 10 \, \text{m/s} \]

3. **Determine the time to reach maximum height:**
At maximum height, the vertical component of the velocity becomes zero. We can use the equation of motion:
\[ v_y = u_y - g t \]
At maximum height, \( v_y = 0 \):
\[ 0 = 10 - 10 t \quad (\text{taking } g = 10 \, \text{m/s}^2) \]
Rearranging gives:
\[ 10 t = 10 \implies t = 1 \, \text{s} \]

4. **Conclusion:**
The time after which the angle between velocity and acceleration will be \( 90^\circ \) is \( 1 \, \text{s} \).

### Final Answer:
The time after which the angle between velocity and acceleration will be \( 90^\circ \) is **1 second**.