Explanation:
Since, the rod is present in vertical position and the spring is unrestrained.
So, initial potential energy stored in the spring is
= 0
And, initial potential gravitational potential energy of the rod is
.
It is given that,
mass of the bar = 0.795 kg
g = 9.8 
L = length of the rod = 0.2 m
Initial total energy T = 
Now, when the rod is in horizontal position then final total energy will be as follows.
T = 
where, I = moment of inertia of the rod about the end = 
Also, 
where,
= speed of the tip of the rod
x = spring extension
The initial unstrained length is
= 0.1 m
Therefore, final length will be calculated as follows.
x' =
m
Then, x = 
x =
m - 0.1 m
= 0.1236 m
k = 25 N/m
So, according to the law of conservation of energy


Putting the given values into the above formula as follows.


v = 2.079 m/s
Thus, we can conclude that tangential speed with which end A strikes the horizontal surface is 2.079 m/s.