A ball, A, is thrown vertically upward from the top of a 30-meter-high building with an initial velocity of 5 m/s. At the same instant, another ball, B, is thrown upward from the ground with an initial velocity of 20 m/s. Determine the height above the ground and the time at which they pass each other.

To solve for the time at which the balls pass each other and the height above the ground at that instant, we'll use the equations of motion for uniformly accelerated motion (since the only acceleration is due to gravity, which we'll assume is \( g = 9.8 \, m/s^2 \) directed downwards).

Let's consider ball A first:

The initial velocity of ball A (u_A) is 5 m/s upwards, and it's thrown from a height of 30 meters (h_A). We can write the equation for the height of ball A at any time t as: \[ h_A(t) = 30 + u_A \cdot t - \frac{1}{2} g t^2 \] \[ h_A(t) = 30 + 5t - 4.9t^2 \]

Now, for ball B:

The initial velocity of ball B (u_B) is 20 m/s upwards, and it's thrown from the ground, which we'll consider as 0 meters (h_B=0). The equation for the height of ball B at any time t is: \[ h_B(t) = u_B \cdot t - \frac{1}{2} g t^2 \] \[ h_B(t) = 20t - 4.9t^2 \]

The balls pass each other when they are at the same height, so we can set \( h_A(t) = h_B(t) \) and solve for t. \[ 30 + 5t - 4.9t^2 = 20t - 4.9t^2 \] Simplifying, we get: \[ 30 = 15t \] \[ t = \frac{30}{15} \] \[ t = 2 \, \text{seconds} \]

Now, let's find out the height at that time using either \( h_A(t) \) or \( h_B(t) \) (since they should give us the same answer). Using \( h_B(t) \): \[ h = 20 \cdot 2 - 4.9 \cdot 2^2 \] \[ h = 40 - 4.9 \cdot 4 \] \[ h = 40 - 19.6 \] \[ h = 20.4 \, \text{meters} \]

So, ball A and ball B pass each other at a height of 20.4 meters above the ground, 2 seconds after they were thrown.

Extra: When dealing with vertical motion under gravity, there are a few key concepts to keep in mind:

- The acceleration due to gravity is constant near the Earth's surface and is directed downwards. In most problems, we approximate this acceleration as \( g = 9.8 \, m/s^2 \). - The initial velocity determines how fast the object starts moving and in which direction it's initially moving. - The equations of motion for uniformly accelerated motion allow us to relate the initial velocity, acceleration, time, and displacement of an object. One key equation is \( s = ut + \frac{1}{2}at^2 \), where s is the displacement, u is the initial velocity, a is the acceleration, and t is time. - In problems involving two objects moving toward each other, like the one with balls A and B, you need to find a common point to equate their positions (heights in this case) to determine when and where they meet or pass each other. - In the absence of air resistance, an object thrown upwards will decelerate due to gravity until its velocity becomes zero at its highest point, after which it accelerates downwards again.