a body of mass 2 kg is thrown vertically upward with kinetic energy 245 J. The acceleration due to gravity is 9.8m/s^2 the kinetic energy of the body will become half at a hieght of
I need a solution
To find the height at which the kinetic energy of the body becomes half, we can use the concept of conservation of mechanical energy. The total mechanical energy (E) of the body is the sum of its kinetic energy (K) and potential energy (PE).
When the body is thrown vertically upward, its initial kinetic energy is given as 245 J. As it rises to a certain height, some of its kinetic energy gets converted into potential energy due to the force of gravity.
At the highest point in its trajectory, the body momentarily comes to rest before falling back down. At this point, all its kinetic energy is converted to potential energy.
Therefore, we can equate the initial kinetic energy to the potential energy at the highest point:
K_initial = PE_highest
Now, let's calculate the height at which the kinetic energy becomes half:
K_initial = 245 J
K_final = 1/2 * K_initial (half of the initial kinetic energy)
Since the body reaches its highest point when its kinetic energy becomes half, we can equate:
PE_highest = K_final
Using the gravitational potential energy formula:
PE = m * g * h
Where:
PE = potential energy
m = mass
g = acceleration due to gravity
h = height
We need to rearrange the formula to solve for h:
h = PE / (m * g)
Now, substitute the known values:
h = K_final / (m * g)
h = (1/2 * K_initial) / (m * g)
h = (1/2 * 245 J) / (2 kg * 9.8 m/s^2)
h = 12.25 J / 19.6 kg.m/s^2
h ≈ 0.625 meters
Therefore, the kinetic energy of the body will become half at a height of approximately 0.625 meters.
total energy launch=total energy later
245J=KE(h) + PE(h)
= 245/2+mgh
h= 245/(2*2*9.8) meters