an object is at tower p. if Velocity at distance h below p is double then at a distance h above p then show that max height above p is 5h/3.

Velocity relative to what?

if an object is at tower p.and then thrown vertically,velocityat distance below h is double then velocity at distance above h.then show max height above p is5h/3.

To prove that the maximum height above point P is 5h/3, where the velocity at a distance h below P is twice as much as the velocity at a distance h above P, we need to use the concepts of projectile motion.

Let's consider the object's motion in two parts: when it is moving downward (below point P) and when it is moving upward (above point P).

1. Downward motion:
When the object is at a distance h below P, its velocity can be considered as v. We can assume that the object is at its highest point of downward motion at this distance.
The formula for the velocity of an object in projectile motion is given by: v^2 = u^2 + 2gh, where u is the initial velocity and g is the acceleration due to gravity.
In this case, the object is at its highest point, so its final velocity (v) is 0. We can rewrite the formula as follows: 0 = u^2 + 2gh.
Since the velocity at a distance h below P is double that at a distance h above P, we have v = 2u for the distance h below P. Plugging this into the equation, we get:
0 = (2u)^2 + 2gh
0 = 4u^2 + 2gh

2. Upward motion:
When the object is at a distance h above P, its velocity can be considered as u. We can assume that the object is at its highest point of upward motion at this distance.
Using the same formula as before, we can write the equation as follows: 0 = u^2 - 2gh.

Now, to find the maximum height above point P, we need to find the difference between the heights at h below P and h above P.
Subtracting the equation for upward motion from the equation for downward motion, we get:
0 = 4u^2 + 2gh - (u^2 - 2gh)
0 = 3u^2 + 4gh

Simplifying the equation, we have:
3u^2 = 4gh

Now, let's solve for u^2:
u^2 = 4gh/3

Since the maximum height occurs when the object reaches its highest point of upward motion (u = 0), we can plug in this value to find the maximum height:
0 = 4gh_max/3

Simplifying the equation, we get:
h_max = 3/4 * (0)
h_max = 0

Therefore, the maximum height above point P is 0.

It seems there was an error in the given statement or the steps followed. The maximum height above point P is not 5h/3.