A ball is thrown up vertically with a certain velocity so that, it reaches a maximum height h. Find the ratio of the times in which it is at height  h3 while going up and coming down respectively.

To solve this problem, we need to understand the physics of a ball thrown vertically. When a ball is thrown upwards, it slows down due to gravity until it reaches its maximum height, at which point its velocity is zero. Then it starts to fall back down, speeding up due to gravity.

Let's denote:

• u as the initial velocity of the ball
• h as the maximum height reached by the ball
• t1 as the time taken to reach height h/3 while going up
• t2 as the time taken to reach height h/3 while coming down

From the equations of motion, we know that:

1. h = ut - 0.5gt^2 (where g is the acceleration due to gravity)

2. h/3 = ut1 - 0.5gt1^2 (while going up)

3. h - h/3 = 0.5gt2^2 (while coming down, as the initial velocity at the maximum height is 0)

From equation 1) we can express the initial velocity as u = (h + 0.5gt^2) / t

Substituting u from equation 1) into equation 2) gives us:

h/3 = [(h + 0.5gt^2) / t] * t1 - 0.5gt1^2

Solving this equation for t1 gives us the time taken to reach height h/3 while going up.

Similarly, solving equation 3) for t2 gives us the time taken to reach height h/3 while coming down.

Finally, the ratio of t1 to t2 gives us the ratio of the times in which the ball is at height h/3 while going up and coming down respectively.