A cricket ball is thrown at a speed of `28 ms^(-1)` in a direction `30^(@)` above the horizontal. Calculate the time taken by the ball to return to the same level.

Explanation

To solve the problem of calculating the time taken by a cricket ball thrown at a speed of \(28 \, \text{ms}^{-1}\) at an angle of \(30^\circ\) above the horizontal to return to the same level, we can follow these steps:

### Step 1: Resolve the initial velocity into horizontal and vertical components.
The initial speed \(v_0\) is given as \(28 \, \text{ms}^{-1}\) and the angle \(\theta\) is \(30^\circ\).

- The vertical component of the velocity \(v_{0y}\) can be calculated using:
\[
v_{0y} = v_0 \sin(\theta) = 28 \sin(30^\circ)
...