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:
500 N
Explanation:
Since the work done on the spring W = Fx where F = force applied and x = compression length = 0.170 m (since the spring will be compressed its full length when the force is applied)
Since W = 85.0 J and we need to find F,
F = W/x
= 85.0 J/0.170 m
= 500 N
So, the magnitude of force must you apply to hold the platform stationary at the final distance given above is 500 N.
Answer:
ok
Explanation:
the correct answer is C please follow me
Answer:
if its so easy why dont u do it .
Explanation:
Hi, it’s the iris. Hope this helped u