**Answer:**

The motion of a ball thrown vertically upwards can be described using the equations of motion.

The equation of motion for vertical motion is given by:

**s = ut - (1/2)gt^2**

Where:
- s is the distance covered
- u is the initial velocity
- t is the time taken
- g is the acceleration due to gravity (approximately 9.8 m/s^2)

Now, we need to find the distance covered during the last second of its ascent. This means we need to find the distance covered when t = (total time - 1).

Let's substitute this value into the equation of motion:

**s = u(total time - 1) - (1/2)g(total time - 1)^2**

To simplify this equation, let's expand the square term:

**s = ut - u - (1/2)gt^2 + gt - (1/2)g**

Combining like terms, we get:

**s = ut - (1/2)gt^2 + gt - (1/2)g - u**

Since we are interested in the distance covered during the last second of ascent, we can substitute t = 1 into the equation:

**s = u(1) - (1/2)g(1)^2 + g(1) - (1/2)g - u**

Simplifying further, we get:

**s = u - (1/2)g + g - (1/2)g - u**

Simplifying the equation, we find that the terms involving u cancel out:

**s = g - (1/2)g - (1/2)g**

Further simplification yields:

**s = 0**

Therefore, the distance covered during the last second of ascent is 0. This makes sense since the ball reaches its maximum height and starts descending during the last second of its ascent, so it doesn't cover any additional distance.

We know,to find distance
S=ut+gt'2/2
g acts downwards -g
S=ut-gt'2/2