Calculating Speeds Of Meeting Balls: Physics Problem

Hey guys! Let's dive into a classic physics problem involving two balls moving under the influence of gravity. This is a common scenario in introductory physics courses, and understanding the concepts here can really solidify your grasp on kinematics. We'll break down the problem step-by-step, making sure to cover all the key principles involved. So, buckle up and let's get started!

Understanding the Problem

Let's first understand the problem deeply. The core of this problem revolves around two balls, P and Q, launched with initial velocities in opposite directions. Ball P is thrown upwards from the ground, while ball Q is thrown downwards from the top of a building. We're given their initial velocities and the height of the building, and our mission is to find the velocities of both balls at the moment they meet. To solve this, we'll need to use our knowledge of kinematics, specifically the equations of motion under constant acceleration (gravity, in this case). Think about it this way: both balls are constantly being pulled downwards by gravity, which affects their speeds as they travel. We need to figure out how this acceleration influences their velocities and positions over time until they finally cross paths. This involves a bit of algebraic manipulation and a good understanding of how displacement, velocity, acceleration, and time are all related in these types of problems. So, before we jump into the calculations, let's make sure we have a clear picture of what's happening physically.

Initial Conditions and Given Values

Before we start crunching numbers, let's clearly outline the given information. This will help us organize our thoughts and choose the right equations. We know:

• Initial velocity of ball P (${v_{P0}}$): 25 m/s upwards
• Initial velocity of ball Q (${v_{Q0}}$): 15 m/s downwards
• Height of the building (h): 80 m
• Acceleration due to gravity (g): We'll assume the standard value of 9.8 m/s² downwards (although the problem implies we might be using 10 m/s², but we'll stick to the more precise value for now and adjust if needed). It's super important to pay close attention to the directions here. Since we're dealing with vertical motion, we need to establish a sign convention. Let's take upwards as positive and downwards as negative. This means the acceleration due to gravity will be -9.8 m/s², and the initial velocity of ball Q will be -15 m/s. Getting these signs right from the beginning is crucial for avoiding errors later on. We also need to keep in mind that the balls meet at the same time and the sum of the distances they travel will equal the building's height. These are the key pieces of the puzzle, and now we're ready to start putting them together to find the solution.

Key Concepts and Equations

To tackle this problem, we need to bring in our trusty tools from kinematics. The main concepts here are uniformly accelerated motion and the equations that describe it. Remember those equations from physics class? They're going to be our best friends here! We'll be using the following equations of motion:

1. Displacement (${Δy}$): ${Δy = v_0t + \frac{1}{2}at^2}$ (where ${v_0}$ is the initial velocity, ${t}$ is the time, and ${a}$ is the acceleration)
2. Final velocity (v): ${v = v_0 + at}$
3. Velocity-displacement relation: ${v^2 = v_0^2 + 2aΔy}$

These equations basically tell us how an object's position and velocity change over time when it's accelerating at a constant rate. In our case, the acceleration is due to gravity. Now, let's think about how these equations apply to our two balls. For each ball, we can write equations for its displacement and velocity as a function of time. We'll have two sets of equations, one for ball P and one for ball Q. The key is to recognize that the time it takes for them to meet is the same for both balls. This shared time will be our link between the two sets of equations, allowing us to solve for the unknowns. We'll also use the fact that the sum of their vertical displacements when they meet equals the total height of the building. By carefully applying these equations and conditions, we'll be able to figure out the velocities of the balls at the moment of their meeting. It's like solving a puzzle where the equations are the pieces, and we need to fit them together to get the complete picture.

Solving the Problem Step-by-Step

Alright, let's get our hands dirty and actually solve this problem! This is where we put the concepts and equations into action. Remember, the key is to break down the problem into smaller, manageable steps. Let's start by defining some variables and setting up our equations. Let's say:

• ${t}$ is the time when the balls meet
• ${y_P}$ is the displacement of ball P when they meet
• ${y_Q}$ is the displacement of ball Q when they meet

Since ball P is thrown upwards from the ground, its initial position is 0. Ball Q is thrown downwards from the top of the building, so its initial position is 80 m. Also, the sum of their displacements should equal the initial height of the building. Now we can write the displacement equations for both balls:

• Ball P: ${y_P = 25t - \frac{1}{2}(9.8)t^2}$
• Ball Q: ${y_Q = -15t - \frac{1}{2}(9.8)t^2}$

Notice the negative sign in front of the gravitational acceleration term – that's because gravity is acting downwards. We also know that the sum of the magnitudes of their displacements must equal the height of the building: ${y_P + |y_Q| = 80}$. Since ${y_Q}$ will be negative (displacement downwards), we'll take the absolute value to represent the distance traveled. Now we have a system of equations that we can solve. We can substitute the expressions for ${y_P}$ and ${y_Q}$ into the distance equation and solve for ${t}$. Once we have the time, we can plug it back into the velocity equations to find the velocities of the balls when they meet. It's a bit of algebra, but if we take it one step at a time, we'll get there! This is where the real problem-solving happens, so let's dive into the math.

Calculating the Time of Meeting

Okay, let's focus on finding the time (${t}$) when the two balls meet. This is a crucial step because once we know the time, we can easily calculate their velocities at that instant. Remember the equations we set up earlier? We have:

• ${y_P = 25t - 4.9t^2}$ (simplified from ${y_P = 25t - \frac{1}{2}(9.8)t^2}$)
• ${y_Q = -15t - 4.9t^2}$ (simplified from ${y_Q = -15t - \frac{1}{2}(9.8)t^2}$)
• ${y_P + |y_Q| = 80}$

Since ${y_Q}$ is negative, we can rewrite the third equation as ${y_P - y_Q = 80}$. Now we can substitute the expressions for ${y_P}$ and ${y_Q}$ into this equation:

${(25t - 4.9t^2) - (-15t - 4.9t^2) = 80}$

Notice that the ${t^2}$ terms cancel out, which makes our life much easier! This simplifies the equation to:

${25t + 15t = 80}$

${40t = 80}$

Now we can easily solve for ${t}$:

${t = \frac{80}{40} = 2}$ seconds

So, the balls meet after 2 seconds! That wasn't so bad, right? The key was to set up the equations correctly and then simplify. Now that we have the time, we're halfway there. The next step is to plug this value back into our velocity equations to find the speeds of the balls when they meet. Let's move on to that now!

Determining the Velocities at Meeting Point

Now for the final piece of the puzzle: calculating the velocities of ball P and ball Q when they meet. We've already found that they meet at ${t = 2}$ seconds. We'll use the equation ${v = v_0 + at}$ to find the final velocities. Remember, we have different initial velocities for each ball, and the acceleration due to gravity is -9.8 m/s². Let's start with ball P:

• Ball P: ${v_P = v_{P0} + at = 25 + (-9.8)(2) = 25 - 19.6 = 5.4}$ m/s

So, the velocity of ball P when they meet is 5.4 m/s upwards (positive value). Now, let's calculate the velocity of ball Q:

• Ball Q: ${v_Q = v_{Q0} + at = -15 + (-9.8)(2) = -15 - 19.6 = -34.6}$ m/s

The velocity of ball Q when they meet is -34.6 m/s, which means 34.6 m/s downwards (negative value). And there you have it! We've successfully calculated the velocities of both balls at the moment they meet. This problem showcases how we can use the equations of motion to analyze the movement of objects under constant acceleration. We took it step by step, from understanding the problem setup to setting up the equations and finally solving for the unknowns. This is the power of physics – breaking down complex scenarios into simpler, manageable components.

Final Answer and Conclusion

Alright guys, we've reached the end of our physics journey for this problem! Let's recap what we found. We were given a scenario with two balls, one thrown upwards and the other thrown downwards, and we needed to find their velocities when they met. After carefully analyzing the problem, setting up our equations, and crunching the numbers, we arrived at the following answers:

• Velocity of ball P at meeting point: 5.4 m/s upwards
• Velocity of ball Q at meeting point: 34.6 m/s downwards

This problem was a great example of how to apply the principles of kinematics to solve real-world scenarios. We used the equations of motion to describe the movement of the balls under the influence of gravity, and we carefully considered the directions of the velocities and acceleration. By breaking the problem down into smaller steps, we were able to tackle it methodically and arrive at the correct solution. Remember, physics isn't about memorizing formulas – it's about understanding the underlying concepts and applying them creatively to solve problems. So, keep practicing, keep exploring, and keep asking questions! You've got this!