To solve the problem, we can break it down using the kinematic equations for vertical motion. Let's start with the key points:

Given:

• Initial velocity in the first case: $v_i$
• Time to reach the top in the first case: $t$
• The new initial velocity in the second case: $2 \times v_i \times a$

Assumptions:

• The acceleration is due to gravity, $g$, and acts downward with a value of $g = 9.8 \, \text{m/s}^2$.
• The ball's velocity at the top of the flight path is zero in both cases.

Step-by-Step Solution:

1. Time to reach the top of the flight path for the first case:

   Using the equation for velocity under constant acceleration:

   $v_f = v_i - g t$

   At the top of the flight path, the final velocity $v_f = 0$, so we can solve for $t$ in terms of $v_i$:

   $0 = v_i - g t$ $t = \frac{v_i}{g}$

2. Time to reach the top of the flight path for the second case:

   In the second case, the initial velocity is $2 \times v_i \times a$, so we can use the same kinematic equation:

   $0 = 2 \times v_i \times a - g t_{\text{new}}$

   Solving for the new time $t_{\text{new}}$:

   $t_{\text{new}} = \frac{2 \times v_i \times a}{g}$

Final Answer:

The time required to reach the top of the flight path when the ball is thrown with an initial velocity of $2 \times v_i \times a$ is:

$t_{\text{new}} = \frac{2 \times v_i \times a}{g}$

Do you want further details or have any questions about this solution?

Here are 5 related questions to explore further:

1. How would the displacement $y_f$ differ in both cases?
2. What would happen to the time if the initial velocity were tripled instead of doubled?
3. How does the value of acceleration $a$ affect the total time in the second case?
4. How would you calculate the total time of flight in each case?
5. What is the impact of air resistance on this motion?

Tip: Remember that when a ball reaches the top of its flight path, its velocity becomes zero for an instant before falling back down.