Note:
The height of a high bar from the floor is h = 2.8 m (or 9.1 ft).
It is not provided in the question, so the standard height is assumed.
g = 9.8 m/s², acceleration due to gravity.
Note that the velocity and distance are measured as positive upward.
Therefore the floor is at a height of h = -2.8 m.
First dismount:
u = 4.0 m/s, initial upward velocity.
Let v = the velocity when the gymnast hits the floor.
Then
v² = u² - 2gh
v² = 16 - 2*9.8*(-2.8) = 70.88
v = 8.42 m/s
Second dismount:
u = -3.0 m/s
v² = (-3.0)² - 2*9.8*(-2.8) = 63.88 m/s
v = 7.99 m/s
The difference in landing velocities is 8.42 - 7.99 = 0.43 m/s.
Answer:
First dismount:
Acceleration = 9.8 m/s² downward
Landing velocity = 8.42 m/s downward
Second dismount:
Acceleration = 9.8 m/s² downward
Landing velocity = 7.99 m/s downward
The landing velocities differ by 0.43 m/s.
Inelastic.
If it was elastic, they'd bump right off each other. But since they've been locked, or stuck together, this is inelastic.
The position function x(t) of a particle moving along an x axis is 
a) The point at which particle stop, it's velocity = 0 m/s
So dx/dt = 0
0 = 0- 12t = -12t
So when time t= 0, velocity = 0 m/s
So the particle is starting from rest.
At t = 0 the particle is (momentarily) stop
b) When t = 0
SO at x = 4m the particle is (momentarily) stop
c) We have
At origin x = 0
Substituting
t = 0.816 seconds or t = - 0.816 seconds
So when t = 0.816 seconds and t = - 0.816 seconds, particle pass through the origin.
In ocean waves, water particles move with mechanical energy and energy moves with gravity
Not sure but hope it helps!
