**Problem Statement:**
A stone is dropped from the top of a tree and it covers a height of 0.2s in the last part of the fall. What is the height of the tree?

**Solution:**
To find the height of the tree, we need to use the equations of motion for free fall under gravity. There are three equations of motion for free fall under gravity:
- v = u + gt
- s = ut + 1/2 gt^2
- v^2 = u^2 + 2gs

where,
- v = final velocity (m/s)
- u = initial velocity (m/s)
- g = acceleration due to gravity (9.8 m/s^2)
- t = time (s)
- s = distance (m)

In this problem, we are given that the stone covers a height of 0.2s in the last part of the fall. Let's assume that the time taken to fall from the top of the tree to the point where it covers a height of 0.2s is t1, and the time taken to fall the remaining height (h-0.2s) is t2.

Then, we can write the following equations:
- h = ut1 + 1/2 gt1^2 + (u + gt1)t2 + 1/2 g t2^2
- 0.2 = ut2 + 1/2 g t2^2

We need to eliminate the variables u and t1 from the above equations. For that, we can use the fact that the initial velocity u is zero, since the stone is dropped from rest. Also, we know that t1 + t2 is the total time taken to fall from the top of the tree to the ground, which is given by t_total = sqrt(2h/g).

Using these facts, we can simplify the above equations as follows:
- h = 1/2 g t_total^2
- 0.2 = 1/2 g t2^2

Solving for t2 from the second equation, we get:
t2 = sqrt(0.4/g)

Substituting this value of t2 in the first equation, we get:
h = 1/2 g (t_total - t2)^2

Substituting the values of g and t_total, we get:
h = 4.9 (sqrt(h/4.9) - sqrt(0.4/4.9))^2

Solving this equation for h, we get:
h = 9.8 m

Therefore, the height of the tree is 9.8 meters.