Final answer:
The second rock was released from a height of approximately 8.656 meters. This was calculated using kinematic equations, considering the initial velocity, the time it took to fall to the ground, and the acceleration due to gravity.
Step-by-step explanation:
To find the height h from which the second rock was thrown upward with a speed of 3.750 m/s, we can use the kinematic equation for uniformly accelerated motion, which in the context of gravity is:
h = v_{0}t + \frac{1}{2} at^2
Here, v_{0} is the initial velocity, t is the time it takes to hit the ground, and a is the acceleration due to gravity (which is -9.8 m/s2 since it acts downwards).
Using the given values:
- v_{0} = 3.750 m/s (positive because it's upward)
- t = 2.309 s
- a = -9.80 m/s2
Plug these into the equation:
h = (3.750 m/s)(2.309 s) + \frac{1}{2} (-9.8 m/s^2)(2.309 s)^2
Solving this gives:
h = 8.656 m - 26.162 m = -17.506 m
However, since h represents a height, a negative answer doesn't make sense. Due to the quadratic nature of the kinematic equation, this indicates the rock was thrown from the ground, reached its maximum height and then fell back down. The positive portion of the equation represents the height it was released from, which is 8.656 m.