To solve this problem, we need to understand the motion of a ball thrown vertically upwards and the kinematics involved in its ascent.

Let's break down the problem:

• The ball is thrown upwards with an initial velocity \(u\).
• We need to find the distance covered during the second-to-last \(t\) seconds of its ascent.
• Let \(T\) be the total time of ascent.

The time taken to reach the topmost point of its trajectory, where its velocity becomes zero, can be calculated using the formula:

\(T = \frac{u}{g}\)

where \(g\) is the acceleration due to gravity.

Now, we are interested in the second-to-last \(t\) seconds of the ascent.

The last \(t\) seconds of the ascent refers to the interval from \(T - t\) to \(T\).

The velocity of the ball at time \(T-t\) can be found using the equation:

\(v = u - g(T-t)\)

Simplifying, using \(T = \frac{u}{g}\), we get:

\(v = u - g \left( \frac{u}{g} - t \right) = gt\)

The velocity of the ball at \(t\) seconds before it reaches the peak is \(gt\).

Now, the distance covered in these \(t\) seconds can be calculated using the equation of motion:

\(s = vt - \frac{1}{2}gt^2\)

Substituting for \(v\) as \(gt\), we have:

\(s = gt \times t - \frac{1}{2}gt^2 = gt^2 - \frac{1}{2}gt^2\)

Simplifying, we get:

\(s = \frac{3}{2}gt^2\)

Thus, the distance covered during the second-to-last \(t\) seconds of its ascent is \(\frac{3}{2}gt^2\).

The correct option is:

$\frac{3}{2}gt^2$

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