A stone is dropped from rest from the roof of a 100-m high building; 2.00 s after that, a second stone is thrown straight down with initial speed v0, and the second stone passes the first stone 21.6 m above ground. (a) When are the stones at the same height? (b) What was the initial speed, v0, of the second stone, in m/s? (c) What are the speeds of the two stones, in m/s, just before they hit the ground?
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To find the answers to these questions, we can use the equations of motion and the principles of kinematics. Let's break down each question step by step:
(a) When are the stones at the same height?
The distance traveled by the first stone can be calculated using the kinematic equation:
s1 = ut + (1/2)at^2,
where s1 is the distance traveled, u is the initial velocity (0 m/s since it is dropped from rest), a is the acceleration due to gravity (-9.8 m/s²), and t is the time.
For the first stone, s1 = -0.5 * 9.8 * t^2 since it is moving downwards.
The distance traveled by the second stone can be calculated through the equation:
s2 = v0t + (1/2)at^2.
Since the second stone is also moving downwards, the equation becomes s2 = -0.5 * 9.8 * t^2 + v0t.
We know that the second stone passes the first stone 21.6 m above the ground, so we have the equation:
-0.5 * 9.8 * t^2 + v0t = -100 + 21.6.
Now, solve this equation to find the value of t when the two stones are at the same height.
(b) What was the initial speed, v0, of the second stone?
After finding the value of t from the previous question, we can use it to solve for the initial speed, v0, of the second stone.
Use the equation:
s2 = v0t + (1/2)at^2,
where s2 is the distance traveled by the second stone.
Substitute the value of t and the given distance (s2 = -100 + 21.6) into this equation to find v0.
(c) What are the speeds of the two stones just before they hit the ground?
To find the speeds of the two stones before they hit the ground, we can use the kinematic equation:
v = u + at,
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
For the first stone, the initial velocity (u1) is 0 m/s, and the acceleration (a) is -9.8 m/s².
For the second stone, use the initial speed (v0) found in question (b) as the initial velocity (u2), and the same acceleration (-9.8 m/s²) for the calculation.
Substitute the values into the equation to find the speeds (v1 and v2) of both stones just before they hit the ground.
By following these steps and using the equations of motion, you should be able to find the answers to all three parts of the question.