Gravitational potential energy i think
Given the equation for the Speed of a Satellite
v = SqRt{Gravitational Constant}{Mass of Earth} divided by the radius given in your problem
we have:
(square root whole term on right side)
v = G Me
———
r
so. (6.67x10^-11)(5.97x10^24)
___________________
(8.0x10^6)
v = 7055 m/s (which is reasonable)
so utilize the Kinetic Energy Formula
KE = 1/2mv^2
KE = 1/2(200)(7055)^2
KE = 4.977x10^9 J
Explanation:
This is true because at maximum height, the velocity is 0
Light that enters the new medium <em>perpendicular to the surface</em> keeps sailing straight through the new medium unrefracted (in the same direction).
Perpendicular to the surface is the "normal" to the surface. So the angle of incidence (angle between the laser and the normal) is zero, and the law of refraction (just like the law of reflection) predicts an angle of zero between the normal and the refracted (or the reflected) beam.
Moral of the story: If you want your laser to keep going in the same direction after it enters the water, or to bounce back in the same direction it came from when it hits the mirror, then shoot it <em>straight on</em> to the surface, perpendicular to it.
Do not worry if you don't recognize both parts of the problem at this point. If you recognize the dynamics problem,<span> On the other hand, if you recognize this as a kinematics problem you will quickly see that you need to find angular acceleration before you can begin and so will need to do that pre-step first.</span>
