00:01
A projectile is launched vertically from the surface of the moon with an initial speed of 1 ,010 meters per second.
00:07
We want to find the altitude when the projectile speed is only one half of its initial value at the surface when it was launched.
00:16
So here i've drawn a diagram of the moon with the red projectile at the surface in position 1 and when it's at 1 half its initial speed at position 2.
00:26
We're going to have to use the conservation of energy equation where the energy at position 1 is equal to the energy at position 2.
00:37
And at position 1 we have kinetic energy as well as gravitational potential energy from the moon and at position 2 we have the terms, the corresponding terms.
00:52
So on the left hand side for kinetic energy we have one half mass of the projectile times the velocity at position 1, which is the initial velocity of 1 ,010 meters per second, minus big g mass of projectile times mass of the moon, divided by a distance, which would be the radius of the moon.
01:14
This would be equal to one half mass of the projectile, the velocity at position 2, minus g m of moon divided by the radius of the moon now plus some altitude or height h...