Answer: The stones are 34.2 meters apart when the second stone has risen for 3.0 seconds.
Step-by-step explanation:
The vertical motion of the stones can be described by the equation:
y = y0 + v0t - 0.5gt^2
where y is the height of the stone, y0 is the initial height (which is 0 for both stones since they are thrown from the ground), v0 is the initial velocity, t is the time, and g is the acceleration due to gravity (9.8 m/s^2).
We can solve for the height of the first stone at t = 3 s and the height of the second stone at t = 4 s, and then subtract these two values to find the distance between the stones.
The height of the first stone at t = 3 s is:
y1 = 0 + v0 * 3 - 0.5 * 9.8 * (3^2)
= 0 + v0 * 3 - 4.9 * 9
= 0 + 3v0 - 44.2
The height of the second stone at t = 4 s is:
y2 = 0 + v0 * 4 - 0.5 * 9.8 * (4^2)
= 0 + v0 * 4 - 4.9 * 16
= 0 + 4v0 - 78.4
The distance between the stones at this time is:
y2 - y1 = (4v0 - 78.4) - (3v0 - 44.2)
= v0 - 34.2
Thus, the stones are 34.2 meters apart when the second stone has risen for 3.0 seconds.