To develop this problem it is necessary to apply the concepts related to Gravitational Potential Energy.
Gravitational potential energy can be defined as
As M=m, then
Where,
m = Mass
G =Gravitational Universal Constant
R = Distance /Radius
PART A) As half its initial value is u'=2u, then
Therefore replacing we have that,
Re-arrange to find v,
Therefore the velocity when the separation has decreased to one-half its initial value is 816m/s
PART B) With a final separation distance of 2r, we have that
Therefore
Therefore the velocity when they are about to collide is 
Answer:
Explanation:
A ) When gymnast is motionless , he is in equilibrium
T = mg
= 63 x 9.81
= 618.03 N
B )
When gymnast climbs up at a constant rate , he is still in equilibrium ie net force acting on it is zero as acceleration is zero.
T = mg
= 618.03 N
C ) If the gymnast climbs up the rope with an upward acceleration of magnitude 0.600 m/s2
Net force on it = T - mg , acting in upward direction
T - mg = m a
T = mg + m a
= m ( g + a )
= 63 ( 9.81 + .6)
= 655.83 N
D ) If the gymnast slides down the rope with a downward acceleration of magnitude 0.600 m/s2
Net force acting in downward direction
mg - T = ma
T = m ( g - a )
= 63 x ( 9.81 - .6 )
= 580.23 N
I believe the acceleration would be 5m/s
All you would need to do is divide the final speed by the time it took to get there. I am only about 80 sure this answer is correct, so take my advise only if you feel comfortable.
A. a tsunami
(if the earthquake is hitting the ocean, the water will get effected)
