a body is thrown upward with velocity v . the height at which P.E is one third of K.E is ?
answer is in terms of v
To find the height at which the potential energy (PE) is one-third of the kinetic energy (KE), we need to understand the relationship between PE and KE.
The potential energy of an object at a certain height is given by the formula PE = mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height.
The kinetic energy of an object is given by the formula KE = (1/2)mv^2, where m is the mass of the object and v is the velocity.
Given that the body is thrown upward with velocity v, we can assume that the final velocity is 0 when it reaches its maximum height. We can use this information to find the height at which PE is one-third of KE.
At the maximum height, the body comes to rest, which means its final velocity is 0. Using the equation of motion v^2 = u^2 - 2gh (where u is the initial velocity), we can substitute u = v and solve for h.
0 = v^2 - 2gh
Rearranging the equation, we get:
2gh = v^2
Now, let's substitute the expressions for PE and KE into the equation:
mgh = (1/2)mv^2
Canceling out the mass, we have:
gh = (1/2)v^2
Since we want to find the height at which PE is one-third of KE, we can write:
h = (1/3)(1/2)v^2
Simplifying this expression, we get:
h = (1/6)v^2
Therefore, the height at which the PE is one-third of KE is (1/6)v^2.