Two stones are thrown vertically up at the same time. The first stone is thrown with an initial velocity of 11.5 m/s from a 12th floor balcony of a building and hits hits the ground after 4.6 seconds. With what initial velocity should the second stone be thrown from a 4th floor balcony so that it hits the ground at the same time as the first stone?

To find the initial velocity of the second stone, we can use the kinematic equation for vertical motion:

h = v_i*t + (1/2)*a*t^2

Where:
h = height (distance traveled by the stone)
v_i = initial velocity
t = time
a = acceleration due to gravity (approximately -9.8 m/s^2)

For the first stone:
h_1 = 12 floors = 12*3.2 m = 38.4 m
v_i_1 = 11.5 m/s
t_1 = 4.6 s

For the second stone:
h_2 = 4 floors = 4*3.2 m = 12.8 m
t_2 = 4.6 s (to hit the ground at the same time as the first stone)

Substituting the values into the equation for each stone:

For the first stone:
38.4 m = 11.5 m/s * 4.6 s - (1/2)*9.8 m/s^2 * (4.6 s)^2

Simplifying:
38.4 m = 52.9 m - 1/2 * 9.8 m/s^2 * 21.16 s^2
38.4 m = 52.9 m - 102.78 m
38.4 m + 102.78 m = 52.9 m
141.18 m = 52.9 m
88.28 m = 52.9 m

For the second stone:
12.8 m = v_i_2 * 4.6 s - (1/2) * 9.8 m/s^2 * (4.6 s)^2

12.8 m = v_i_2 * 4.6 s - 1/2 * 9.8 m/s^2 * 21.16 s^2
12.8 m + 1/2 * 9.8 m/s^2 * 21.16 s^2 = v_i_2 * 4.6 s
12.8 m + 51.67 m = v_i_2 * 4.6 s
64.47 m = v_i_2 * 4.6 s

Dividing both sides by 4.6 s:
64.47 m / 4.6 s = v_i_2

Therefore, the initial velocity of the second stone should be approximately 14.03 m/s to hit the ground at the same time as the first stone.

To find the initial velocity of the second stone, we need to understand the relationship between time, velocity, and displacement for an object thrown vertically upwards.

When a stone is thrown vertically upwards, its initial velocity decreases due to the force of gravity until it reaches its highest point, where its velocity becomes zero. After that point, the stone accelerates downwards under the force of gravity.

We can use the equation for displacement to determine the relationship between time, velocity, and displacement:

s = ut + (1/2)at^2

Where:
s is the displacement
u is the initial velocity
t is the time
a is the acceleration (acceleration due to gravity, which is approximately -9.8 m/s² for objects near the Earth's surface)

For both stones, the final displacement is equal to the height of the building, which is the difference in floor levels between the balcony and the ground. In this case, it is (12 - 1) = 11 floors for the first stone and (4 - 1) = 3 floors for the second stone.

Here's the calculation:

For the first stone:
s1 = 11 floors = 11 * 3 meters (each floor is assumed to be 3 meters)
t1 = 4.6 seconds

Using the equation, we can rearrange it to solve for initial velocity (u):

s1 = ut1 + (1/2)a(t1)^2

11 * 3 = u * 4.6 + (1/2) * (-9.8) * (4.6)^2

33 = 4.6u - 10.678

4.6u = 43.678

u = 43.678 / 4.6

u ≈ 9.5 m/s

So, the initial velocity of the first stone is approximately 9.5 m/s.

Now, let's find the initial velocity for the second stone:

s2 = 3 floors = 3 * 3 meters

Using the same equation as before, but substituting the values for the second stone:

s2 = ut2 + (1/2)a(t2)^2

9 = u * t2 + (1/2) * (-9.8) * (t2)^2

As we want both stones to hit the ground at the same time, we can set t2 = 4.6 seconds, which is the time it takes for the first stone to hit the ground.

9 = u * 4.6 + (1/2) * (-9.8) * (4.6)^2

9 = u * 4.6 + (-4.9) * (4.6)^2

9 = 4.6u - 49.176

4.6u = 58.176

u ≈ 12.7 m/s

Therefore, the second stone should be thrown with an initial velocity of approximately 12.7 m/s from the 4th floor balcony to hit the ground at the same time as the first stone.