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 the ground after 4.0 s. 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? Assume equal height floors, and that in each case the stone is dropped from the same height as the ceiling.

Question

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 the ground after 4.0 s. 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? Assume equal height floors, and that in each case the stone is dropped from the same height as the ceiling.

Answer 1

To determine the initial velocity of the second stone, we need to consider the motion equations used to describe the vertical motion of objects under constant acceleration.

Let's start by identifying the known values from the problem:

For the first stone:
- Initial velocity (u1) = 11.5 m/s (upward)
- Time (t1) = 4.0 s
- Displacement (s1) = height of the building = 12 floors

For the second stone:
- Initial velocity (u2) = ? (upward)
- Time (t2) = 4.0 s
- Displacement (s2) = height of the building = 4 floors

Using the motion equation s = ut + (1/2)at^2, where s is the displacement, u is the initial velocity, t is the time, and a is the acceleration, we can solve for the acceleration (a) for both stones.

For the first stone:
s1 = u1t1 + (1/2)at1^2
12 floors = (11.5 m/s)(4.0 s) + (1/2)a(4.0 s)^2

For the second stone:
s2 = u2t2 + (1/2)at2^2
4 floors = u2(4.0 s) + (1/2)a(4.0 s)^2

Since both stones experience the same acceleration due to gravity (g ≈ 9.8 m/s^2) and the time is the same for both, we can set up the following equation:

12 = 11.5(4) + (1/2)(9.8)(4)^2
4 = u2(4) + (1/2)(9.8)(4)^2

Now we can solve these two equations to find the initial velocity (u2) of the second stone:

12 = 46 + 78.4
4 = 4u2 + 78.4

Rearranging the second equation:
4u2 = 4 - 78.4
4u2 = -74.4
u2 = -74.4 / 4
u2 = -18.6 m/s

Since velocity is a vector quantity and given that both stones are thrown upward, the negative sign indicates the direction of motion. However, for the purpose of this question, we are only interested in the magnitude of the initial velocity. Hence, the initial velocity of the second stone should be 18.6 m/s.

Therefore, to achieve the same time of impact, the second stone needs to be thrown upward with an initial velocity of 18.6 m/s from the 4th-floor balcony.