Calculating Initial Velocity from Time of Flight

The problem asks for the initial velocity (\(V_0\)) of a ball thrown upwards, given the total time it takes to return to the starting point.

Physics Principles for Vertical Motion

For an object thrown vertically upwards and returning to the same point:

• The time taken to reach the maximum height is equal to the time taken to fall back from the maximum height to the starting point.
• The total time of flight (\(T\)) is the sum of the time going up (\(t_{up}\)) and the time coming down (\(t_{down}\)). Thus, \(T = t_{up} + t_{down}\).
• At the maximum height, the ball's instantaneous velocity is 0.
• Acceleration due to gravity (\(g\)) acts downwards, opposing the initial upward motion. We'll use \(g \approx 9.8 \, \text{m/s}^2\).

Step-by-Step Calculation

1. Determine Time to Maximum Height: The total time of flight is given as \(T = 6\) seconds. Since the time going up equals the time coming down, the time to reach the maximum height is half the total time:

   \(t_{up} = \frac{T}{2} = \frac{6 \, \text{s}}{2} = 3 \, \text{s}\)

2. Apply Kinematic Equation: We use the first equation of motion: \(v = V_0 + at\). Here, \(v\) is the final velocity, \(V_0\) is the initial velocity, \(a\) is the acceleration, and \(t\) is the time. At the maximum height, the final velocity (\(v\)) is 0. The acceleration (\(a\)) is \(-g\) (negative because gravity opposes the upward motion). The time (\(t\)) is \(t_{up}\).

   \(0 = V_0 + (-g) \times t_{up}\)

   \(0 = V_0 - g \times t_{up}\)

3. Solve for Initial Velocity (\(V_0\)): Rearrange the equation to solve for \(V_0\):

   \(V_0 = g \times t_{up}\)

4. Substitute Values and Calculate: Substitute the values for \(g\) and \(t_{up}\):

   \(V_0 = (9.8 \, \text{m/s}^2) \times (3 \, \text{s})\)

   \(V_0 = 29.4 \, \text{m/s}\)

Conclusion

The initial velocity with which the ball was thrown is 29.4 m/s.