The problem you have shared involves a projectile motion where a body is projected with an initial velocity of 20 m/s at an angle of $30^\circ$ with the horizontal. We need to find the time it takes for the body to reach its maximum height. The gravitational acceleration is given as $10 \, \text{m/s}^2$.

Step-by-step solution:

1. Vertical Component of the Initial Velocity: The vertical component of the initial velocity ($v_{y0}$) can be calculated using the formula: $v_{y0} = v_0 \sin \theta$ where:

   • $v_0 = 20 \, \text{m/s}$ is the initial velocity,
   • $\theta = 30^\circ$.

   Therefore, $v_{y0} = 20 \sin 30^\circ = 20 \times 0.5 = 10 \, \text{m/s}.$

2. Time to Reach Maximum Height: At the maximum height, the vertical velocity becomes zero. We can use the following equation to find the time ($t$) to reach this point: $v_y = v_{y0} - g t$ At maximum height, $v_y = 0$, so: $0 = 10 - 10t$ Solving for $t$: $t = \frac{10}{10} = 1 \, \text{second}.$

Final Answer:

The time taken by the body to reach the maximum height is 1 second (option (a)).

Let me know if you would like further details or have any questions. Here are five related questions that expand on this topic:

1. How would the time to reach maximum height change if the projection angle is increased?
2. What is the maximum height reached by the body in this scenario?
3. How would the total time of flight be calculated for this projectile?
4. How does air resistance affect the time to reach the maximum height?
5. How can we determine the horizontal range of the projectile?

Tip: In projectile motion, the vertical and horizontal components are independent of each other; always consider them separately for accurate results.