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?

T fast =3t slow

9=3t
T=9/3

slow stone 3s

hight

To solve these problems, we can use the equations of motion for an object in free fall.

Let's assume the initial speed of the slower stone is v and the initial speed of the faster stone is 3v.

1. How long will it take the slower stone to return?
The time it takes for an object to return to the ground when thrown vertically upward is twice the time it takes to reach its maximum height. So, if we can find the time it takes for the slower stone to reach its maximum height, we can double that to find the total time.

Using the equation of motion for time:
v = u + gt
where v is the final velocity, u is the initial velocity, g is the acceleration due to gravity, and t is the time.

For the slower stone:
0 = v + (-9.8)t (since the final velocity is 0 at the maximum height)
0 = u + (-9.8)t (substituting v = u)

We can solve this equation to find the time it takes for the slower stone to reach its maximum height:
0 = vt - 4.9t^2
t(9.8t - v) = 0

Since time cannot be negative, t = 0 or 9.8t - v = 0
Since we're looking for the time it takes to reach the maximum height, we can ignore t = 0.
So, 9.8t - v = 0
t = v/9.8

Doubling this time will give the total time it takes for the slower stone to return:
Total time = 2 * (v/9.8) = 2v/9.8 = 0.2v

Therefore, it will take 0.2v seconds for the slower stone to return.

2. How high will the faster stone go?
To find the maximum height reached by an object thrown vertically upward, we can use the equation:
v^2 = u^2 + 2gh
where h is the maximum height.

For the slower stone:
0 = (v^2) - (3v^2) + 2(-9.8)h
-2v^2 = -19.6h
h = v^2/2g

For the faster stone:
0 = (9v^2) - (3v^2) + 2(-9.8)h'
6v^2 = 19.6h'
h' = 6v^2/2g

Comparing the two equations, we can see that the faster stone will reach a maximum height that is 6 times higher than the slower stone:
h' = 6 * h

Therefore, the faster stone will go to a maximum height that is 6 times higher than the slower stone, or 6h.

To solve this problem, we can use the equations of motion for objects in free fall.

Let's denote the initial speed of the slower stone as v and the initial speed of the faster stone as 3v (since it's three times faster).

1. Finding the time of flight for the slower stone:
The time it takes for an object thrown vertically upward to return to the ground is twice the time it takes to reach the maximum height. So, we need to find the time it takes for the slower stone to reach its maximum height.

Using the first equation of motion, we know that the final velocity at the maximum height is 0 m/s. The initial velocity is v, and the acceleration due to gravity, assuming free fall, is -9.8 m/s^2 (negative because it acts in the opposite direction to the initial velocity).

Using the equation:
v_f = v_i + at

0 = v + (-9.8)t_max

Rearranging the equation, we get:
t_max = v/9.8

2. Finding the time of flight for the faster stone:
Given that the faster stone takes 9.00 seconds to return to the ground, we can use the same reasoning as before. The time for the faster stone to reach its maximum height will be half of 9.00 seconds.

So, t_max_fast = 9.00/2 = 4.50 seconds.

3. Finding the maximum height reached by the faster stone:
Using the second equation of motion, we can find the maximum height reached by the faster stone.

The final velocity at the maximum height is 0 m/s. The initial velocity is 3v, and the acceleration due to gravity is -9.8 m/s^2.

Using the equation:
v_f^2 = v_i^2 + 2ad

0 = (3v)^2 + 2(-9.8)d_max

Simplifying the equation, we get:
d_max = (9v^2)/(2 * 9.8)

So, the maximum height reached by the faster stone is (9v^2)/(19.6) meters.

To find the answer to the second question, we need the value of the maximum height of the slower stone (which is missing in your question). Without that value, we cannot evaluate the height of the faster stone.