To determine the height at which the potential energy becomes double the kinetic energy, we need to equate the expressions for potential energy and kinetic energy and solve for the height.
Given: Mass of the object (m) = 4 kg Initial velocity (v) = 40 m/s
The potential energy (PE) of an object at a certain height is given by: PE = mgh
The kinetic energy (KE) of an object is given by: KE = (1/2)mv^2
At the desired height, we want the potential energy to be double the kinetic energy. Mathematically, we can write:
PE = 2KE
Substituting the expressions for PE and KE:
mgh = 2[(1/2)mv^2]
Canceling out the common terms:
gh = v^2
Now, let's solve for h, the height:
h = (v^2) / g
Substituting the given values: v = 40 m/s g = 9.8 m/s^2 (acceleration due to gravity)
h = (40^2) / 9.8
h ≈ 163.27 meters
Therefore, at a height of approximately 163.27 meters, the potential energy will be double the kinetic energy for the given object thrown upward with a mass of 4 kg and an initial velocity of 40 m/s.