00:01
In this problem, we have to find out the maximum height above the surface of earth for an object which is launched upward at 5 .1 km per second from the surface.
00:15
So, here we will use the energy conservation.
00:22
Energy at the surface, initial energy, is equal to energy at maximum height, which is the final energy.
00:31
Now at the surface we have potential energy plus initial kinetic energy and at the maximum height the speed will be zero so we will only have final potential energy.
00:51
Now initial potential energy at the surface is minus gmm upon ri where ri is the initial separation from the center of earth and initial.
01:05
The object is at surface, so the distance from the center will be equal to radius of earth.
01:14
Capital m is mass of earth and small m is mass of the object.
01:18
G is gravitational constant.
01:21
Kinetic energy is half m vi square.
01:26
Vi is initial speed with which it is launched...