Let's solve the problem step-by-step.

When a ball is thrown vertically upwards with an initial velocity $v_0$, we know:

• The ball reaches its maximum height when its velocity becomes zero.
• The downward acceleration due to gravity is $g$.

The time taken to reach the maximum height is given by:

$t = \frac{v_0}{g}$

Since the motion is symmetrical, the time to return to the starting point (falling back down) is also $\frac{v_0}{g}$. Therefore, the total time for the upward and downward journey is:

$T = 2 \times \frac{v_0}{g} = \frac{2v_0}{g}$

Hence, the correct answer is:

(D) $\frac{2v_0}{g}$

Do you need further clarification or detailed explanations?

Related Questions:

1. How do you derive the time for a ball to reach maximum height when thrown upward?
2. What is the equation for the maximum height reached by a projectile?
3. What factors affect the total time of flight for a vertically thrown object?
4. How do we calculate the velocity of an object just before it hits the ground?
5. How does air resistance affect the total time of flight?

Tip:

For vertical motion problems, always remember that the time to go up equals the time to come down when ignoring air resistance.