Let h = distance (m) to the water surface.
Initial velocity, u = 0 (because the stone was dropped).
Use the formula
h = ut + (1/2)gt^2
where g = 9.8 m/s^2 (acc. due to graity)
t = time (s)
h = (1/2)*(9.8)*(3^2) = 44.1 m
If a coin is dropped at a relatively low altitude, it's acceleration remains constant. However, if the coin is dropped at a very high altitude, air resistance will have a significant effect. The initial acceleration of the coin will be the greatest. As it falls down, air resistance will counteract the weight of the coin. So, the acceleration will decrease. Although the acceleration decreases, the coin still accelerates, that is why it falls faster. When the air resistance fully counters the weight of the coin, the acceleration will become zero and the coin will fall at a constant speed (terminal velocity). So, the answer should be, The acceleration decreases until it reaches 0. The closest answer is.
a. The acceleration decreases.
Resultant force= (2*6^2)^(1/2)
=8.5m/s
answer is B.
22.5 J
Explanation:
Given:
x = 3 m
The spring potential energy 
is
Answer:2155 J
Explanation:
Given
Change in Internal energy 
i.e. decrease in Internal Energy
Heat added to system 
According First law for a system
Thus 2155 J of work is done by system
