The body passes a point at h height above twice, once when it's going upward & again while coming downwardu = 52 m/sDistance travelled in the interval of 10 secondss = ut + 1/2 gt^2s = (52×10) + 1/2×(-10)×10^2s = 520 - 500s = 20 mAs it passes twice a point of height h, we can write:s = 2h = 20 mh = 10 m

Analysis of the Problem:
The given problem involves a body being thrown vertically up with an initial velocity of 52 m/s and passing twice a point at a height h above the ground at an interval of 10 s. We need to determine the height h of the point.

Key Points:
- Initial velocity of the body: 52 m/s
- Time interval between passing the point twice: 10 s

Calculating the Height:
1. The motion of the body can be analyzed using the kinematic equation: h = v0t + (1/2)gt^2, where:
- h is the height of the body above the ground
- v0 is the initial velocity of the body
- t is the time elapsed
- g is the acceleration due to gravity (9.8 m/s^2)
2. The body passes the point at h height above the ground twice, which means it reaches the same height at two different instances. Let's denote the two instances as t1 and t2.
3. At time t1, the height of the body is h:
h = v0t1 - (1/2)gt1^2
4. At time t2, the height of the body is h:
h = v0t2 - (1/2)gt2^2
5. The time interval between passing the point twice is given as 10 s:
t2 - t1 = 10 s
6. By substituting the values of h from equations (3) and (4) into equation (5), we can solve for the height h.
7. Solving the equations will give us the height h of the point above the ground.
Therefore, by following the above steps and solving the equations, we can determine the height h of the point above the ground.