The second stone hits the ground exactly one second after the first.
The distance traveled by each stone down the cliff is calculated using second kinematic equation;

where;
- <em>t is the time of motion </em>
- <em />
<em> is the initial vertical velocity of the stone = 0</em>

The time taken by the first stone to hit the ground is calculated as;

When compared to the first stone, the time taken by the second stone to hit the ground after 1 second it was released is calculated as


Thus, we can conclude that the second stone hits the ground exactly one second after the first.
"<em>Your question is not complete, it seems be missing the following information;"</em>
A. The second stone hits the ground exactly one second after the first.
B. The second stone hits the ground less than one second after the first
C. The second stone hits the ground more than one second after the first.
D. The second stone hits the ground at the same time as the first.
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Answer:
The correct answer to the question is
B. It always decreases
Explanation:
To solve the question, we note that the foce of gravity is given by
where
G= Gravitational constant
m₁ = mass of first object
m₂ = mass of second object
r = the distance between both objects
If the mass of one object remains unchanged while the distance to the second object and the second object’s mass are both doubled, we have
= 
Therefore the gravitational force is halved. That is it will always decrease
T = 3.5 secs
Velocity (v) = g * t = 10 m/s^2 * 3.5 sec = 35 m/s
Answer:
(A) We are using them faster than they are replenished by nature
There isnt enough information to answer the question, the missing variable is "distance from said falling spot and ground"