Answer:
Explanation:
Given
mass of block 
at 
displacement is 
velocity 
acceleration 
suppose
is the general equation of SHM
where A=amplitude
=natural frequency of oscillation
therefore velocity and acceleration is given by


for t=0



divide 1 and 3 we get


Now square and 1 and 2 we get



Answer:
Well, I think you're talking about kinematics, especially uniform rectilinear motion. We know that there is a specific equation for that:
S = Vt + S0
With S being the distance, V the velocity, t the time and S0 the initial distance (initial displacement).
From this you can calculate t, if that's what you want.
I assume that the ball is stationary (v=0) at point B, so its total energy is just potential energy, and it is equal to 7.35 J.
At point A, all this energy has converted into kinetic energy, which is:

And since K=7.35 J, we can find the velocity, v:
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