Explanation:

When an object falls freely under the influence of gravity, its speed increases due to the constant acceleration provided by gravity.

The acceleration due to gravity is denoted by 'g' and is approximately equal to 9.8 m/s^2 on the surface of the Earth.

Therefore, in the first second of free fall, the object falls a distance of h/2. In the second second, it falls a distance of 3h/2.

Answer:

The fall of height in the next second would be h+g.

Explanation:

In the first second, the object falls a distance of h/2. In the second second, it falls a distance of 3h/2. Therefore, the total distance fallen in the first two seconds is:

h/2 + 3h/2 = 2h

This means that the average speed of the object during the first two seconds is:

average speed = total distance / total time
= 2h / 2s
= h/s

During the third second, the object will continue to accelerate under the influence of gravity. Its final velocity at the end of the second second will be:

vf = vi + at
= 0 + g(2s)
= 2gs

where vi is the initial velocity (which is zero), t is the time (which is 2s), and a is the acceleration due to gravity (which is g).

At the beginning of the third second, the object will be moving with a velocity of 2gs. It will continue to accelerate at a rate of g for the entire third second. Therefore, the distance it falls during the third second will be:

d = vi*t + 1/2 * a * t^2
= 2gs * 1s + 1/2 * g * 1s^2
= 2gs + 1/2 * g

Adding this distance to the total distance fallen in the first two seconds gives:

2h + 2gs + 1/2 * g

Simplifying this expression gives:

h + g/2 + 2gs

But we know that 2gs = 2g * 1s = 2g, since the object has been falling for two seconds. Therefore, the total distance fallen in the first three seconds is:

h + g/2 + 2g

Simplifying this expression gives:

h + 5g/2

Therefore, the fall of height in the next second would be:

(h + 5g/2) - (h + g/2) = 4g/2 = 2g

Adding this to the distance fallen in the first two seconds gives:

2h + 2gs + 2g = h + 5g

Simplifying this expression gives:

h + g(5)

Therefore, the fall of height in the next second would be h+g.

The Answer will be (c)h+g ,as from given information we obtained relation between h and nORt(time)as, -. h=g(2t-1)/2 ....... (equation no. 1) and for (n+1)th time let the hight covered is H ,,,,so H=g{2(n+1)-1}\2={g(2n-1)/2}+g,,, then from eqtn(1) we get, H=(h+g)