If a ball is thrown vertically upwards with speed u, the distance covered during the last t seconds of its ascent is

To find the distance covered by the ball during the last $t$ seconds of its ascent, we can analyze the motion of the ball in reverse.

When a ball is thrown vertically upwards with an initial speed $u$, it decelerates under gravity $g$ until it reaches its highest point (peak), where its velocity becomes zero.

By the principle of time symmetry in vertical motion under gravity (neglecting air resistance), the upward journey of the ball during its last $t$ seconds of ascent is identical to the downward journey during the first $t$ seconds of its descent from the highest point.

At the highest point, the initial velocity for the descent is:
$v=0$

Using the second equation of motion for the downward descent:
$s=ut+\frac{1}{2}a{t}^2$
Here, the initial velocity is $0$, and the acceleration $a$ is equal to $g$.

Substituting these values:
$s=0\cdot t+\frac{1}{2}g{t}^2$
$s=\frac{1}{2}g{t}^2$

Therefore, the distance covered during the last $t$ seconds of its ascent is equal to the distance covered in the first $t$ seconds of free fall from rest, which is $\frac{1}{2}g{t}^2$.

The correct option is (1/2)gt².