A ball is projected so as to reach a height h passes points p and q with velocites v/2 and v/3 where v is the initial velocity with which the body is thrown the distance between p and q in terms of h is

To find the distance between points p and q in terms of height (h), we need to analyze the motion of the ball.

Let's break down the problem into two parts:

1. Vertical Motion:
Since the ball reaches a height h, we can use the equations of motion to determine the time it takes to reach that height.
The equation relating the initial velocity (u), final velocity (v), acceleration (a), and time (t) for an object in vertical motion is:
v^2 = u^2 + 2ah

In this case, the initial velocity is v, and the final velocity is 0 (when the ball reaches its maximum height). Plugging in these values, we get:
0 = v^2 - 2ah

Solving for time (t), we have:
t = v / (2 * sqrt(h))

2. Horizontal Motion:
The ball passes through points p and q with velocities v/2 and v/3, respectively. Considering the horizontal motion, we know that the horizontal component of velocity remains constant throughout the motion. Therefore, the ratio of the distances covered by the ball when passing through points p and q is the same as the ratio of the corresponding velocities.

Let's denote the distance between points p and q as d.

Using the ratio of velocities:
(v/2) / (v/3) = d / (d - h)

Simplifying this equation:
3 / 2 = d / (d - h)

Cross-multiplying and rearranging, we get:
2d - 2h = 3d
d = 2h

Therefore, the distance between points p and q (d) in terms of height (h) is 2h.