body is projected vertically up. At certain height h above the ground its PE and KE are in the ratio 1:4. At what height above the ground PE and KE will be in the ratio 4:1?

To solve this problem, let's assume that the initial height of the body above the ground is h1 and the final height is h2. We are given that at a certain height h above the ground, the potential energy (PE) and kinetic energy (KE) are in the ratio 1:4.

We can use the formulas for potential energy and kinetic energy to set up an equation. The potential energy (PE) is given by PE = mgh, where m is the mass of the body, g is the acceleration due to gravity, and h is the height above the ground. The kinetic energy (KE) is given by KE = (1/2)mv^2, where v is the velocity of the body.

Since the mass and velocity of the body remain constant, we can write the ratio of PE to KE as (mgh)/(1/2)mv^2 = 1/4. Simplifying this equation, we get gh/v^2 = 1/8.

Now, let's consider the height h2 above the ground where the PE and KE are in the ratio 4:1. Using the same formulas, we can write the ratio of PE to KE as (mgh2)/(1/2)mv^2 = 4/1. Simplifying this equation, we get gh2/v^2 = 8/1.

To find the height h2, we can set up a proportion using the ratios we derived earlier. We have (gh2/v^2)/(gh/v^2) = (8/1)/(1/8). Simplifying this proportion, we get h2/h = 64/1.

Therefore, the height h2 above the ground where the PE and KE will be in the ratio 4:1 is 64 times the initial height h.