Two stones are thrown vertically upward from the ground, one with three times the initial speed of the other. Assume free fall. If the faster stone takes 9.00 to return to the ground, how long will it take the slower stone to return? If the slower stone reaches a maximum height of , how high (in terms of ) will the faster stone go?

first one is 3

To find the time it takes for the slower stone to return to the ground, we can use the fact that the total time for the stone to go up and then come back down is twice the time it takes for the stone to reach its maximum height.

Let's call the initial speed of the slower stone v and the time it takes for it to reach its maximum height t.

For the slower stone:
Upward velocity = v
Time to reach maximum height = t
Time to return to the ground = 2t

Next, let's consider the faster stone. We are told that it has three times the initial speed of the slower stone, so its initial speed is 3v.

For the faster stone:
Upward velocity = 3v
Time to reach maximum height = t'
Time to return to the ground = 2t'

We are given that the faster stone takes 9.00 seconds to return to the ground, so we can set up the equation:

2t' = 9.00

Solving for t', we find:

t' = 4.50 seconds

To find the time it takes for the slower stone to return, we can substitute t' = 4.50 seconds into the equation for the faster stone:

2t = 2(4.50) = 9.00 seconds

So, it will take the slower stone 9.00 seconds to return to the ground.

To find how high the faster stone will go, we need to find its maximum height.

The equation for calculating the maximum height of an object thrown vertically upward is:

Maximum height = (initial velocity^2) / (2 * acceleration due to gravity)

For the faster stone:
Initial velocity = 3v
Acceleration due to gravity = -9.8 m/s^2 (assuming downward is positive)

Maximum height = (3v)^2 / (2 * 9.8) = 9v^2 / 19.6

Therefore, the faster stone will reach a maximum height of 9v^2 / 19.6.

To solve this problem, we can use the kinematic equations of motion for objects in free fall. Let's break down each part of the problem separately:

1. The time taken for the faster stone to return to the ground:
We are given that the faster stone takes 9.00 seconds to return to the ground. Since the stone is thrown vertically upward and it takes the same amount of time to reach the maximum height and return back to the ground, we can say that the time taken for the upward journey is half of the total time. Therefore, the time taken for the upward journey of the faster stone is 9.00/2 = 4.50 seconds.

2. The time taken for the slower stone to return to the ground:
Since the slower stone is thrown with one-third the initial speed of the faster stone, we can use the concept of proportionalities. The time taken for an object to reach its maximum height and return to the ground is directly proportional to its initial velocity. So, if the initial velocity is one-third of the faster stone's initial velocity, the time taken by the slower stone will also be one-third of the time taken by the faster stone. Hence, the time taken for the slower stone to return to the ground is 4.50 * 1/3 = 1.50 seconds.

3. The maximum height reached by the faster stone:
To find the maximum height reached by the faster stone, we can use the equation of motion:
vf^2 = vi^2 + 2as
where:
vf is the final velocity (0 m/s at the maximum height),
vi is the initial velocity,
a is the acceleration (which is equal to -9.8 m/s^2 for objects in free fall), and
s is the displacement (which is the maximum height).
Rearranging the equation, we get:
s = (vf^2 - vi^2) / (2a)
Since the final velocity at the maximum height is 0 m/s, we can simplify the equation to:
s = -vi^2 / (2a)
Substituting the initial velocity of the faster stone (3 times the initial velocity of the slower stone) into the equation, we get:
s = -(3vi)^2 / (2a)
Simplifying further, we have:
s = -9vi^2 / (2a)
Now, using the relationship between the initial velocity and the time of flight, we know that the time of flight (4.50 seconds) is equal to 2vi / a. Rearranging the equation, we can express the initial velocity (vi) as:
vi = a * t / 2
Substituting this value into the equation for the maximum height, we get:
s = -9(a * t / 2)^2 / (2a)
Simplifying further, we have:
s = -81at^2 / 8
So, the maximum height reached by the faster stone is -81at^2 / 8.

Note: The displacement value, s, is negative because it represents a height above the ground, and upward directions are taken as positive.

Therefore, the slower stone takes 1.50 seconds to return to the ground, and the faster stone reaches a maximum height of -81at^2 / 8.