Projectile Motion Analysis Determining Time And Relative Velocity

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This article delves into the fascinating world of projectile motion, specifically examining the scenario of two balls thrown vertically into the air with identical initial velocities but at different times. We aim to determine the time difference, denoted as t1, between the launches of the two balls, as well as the relative velocity of the second ball with respect to the first ball at the moment they pass each other in mid-air. This exploration requires a solid understanding of kinematics principles, including the effects of gravity on projectile motion and the concept of relative motion. The problem combines both conceptual understanding and mathematical application, making it an excellent exercise in physics problem-solving.

Understanding Projectile Motion

To effectively analyze the motion of the balls, it is crucial to first establish a firm grasp on the fundamental concepts of projectile motion. Projectile motion is a type of motion where an object is thrown near the Earth's surface and moves along a curved path under the action of gravity alone. The key assumptions we make in analyzing projectile motion are that air resistance is negligible and that the acceleration due to gravity, denoted as g, is constant and acts vertically downwards. The motion can be conveniently separated into two independent components: horizontal and vertical. In this specific case, since the balls are thrown vertically, we are primarily concerned with the vertical component of the motion. The vertical motion is uniformly accelerated motion under the influence of gravity, which means the velocity changes at a constant rate. The acceleration due to gravity is approximately 9.8 m/s², and it causes objects to slow down as they move upwards and speed up as they fall downwards. The initial upward velocity of the ball will decrease due to gravity until it momentarily becomes zero at the highest point of the trajectory. After reaching this point, the ball begins to descend, with its velocity increasing in the downward direction. By analyzing these principles, we can use kinematic equations to model and predict the behavior of the balls.

Problem Setup and Key Equations

Before diving into the calculations, let's clearly define the problem and identify the relevant kinematic equations. The problem states that two balls are thrown vertically upwards with an initial velocity of 44 m/s. The second ball is thrown t1 seconds after the first ball. We need to find t1 and the relative velocity of the second ball with respect to the first ball when they pass each other. To solve this, we will use the following kinematic equations which describe the motion of objects under constant acceleration:

  1. v = u + at (where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time)
  2. s = ut + (1/2)at² (where s is the displacement)
  3. v² = u² + 2as

In our scenario, u = 44 m/s, a = -9.8 m/s² (negative because gravity acts downwards), and we will use these equations to describe the motion of each ball. It's also essential to recognize that when the balls pass each other, they will be at the same vertical displacement from the ground. This condition will be crucial in setting up our equations to solve for t1. By carefully applying these equations and considering the timing difference in the launches, we can determine both t1 and the relative velocity.

Determining the Time t1

To determine the time t1, we need to analyze the motion of both balls simultaneously and consider the condition that they meet at the same vertical position. Let's denote the time elapsed since the first ball was thrown as t. Therefore, the time elapsed since the second ball was thrown would be t - t1. Let s represent the vertical displacement at which the balls meet. For the first ball, the displacement can be expressed using the kinematic equation:

s = 44t - (1/2)(9.8)t²

For the second ball, the displacement can be expressed as:

s = 44(t - t1) - (1/2)(9.8)(t - t1)²

Since both balls are at the same vertical position when they meet, we can equate the two expressions for s:

44t - (1/2)(9.8)t² = 44(t - t1) - (1/2)(9.8)(t - t1)²

Simplifying this equation involves expanding the terms and canceling out common terms on both sides. This algebraic manipulation will lead us to a relationship that allows us to solve for t1. The key here is to carefully expand the squared terms and collect like terms. After simplification, we will obtain an equation that relates t, t1, and constants. However, we still need another condition to solve for t1 uniquely. That condition arises from understanding that the second ball is thrown while the first ball is still in the air. The maximum height the first ball reaches can be found using the kinematic equation v² = u² + 2as. At the maximum height, the final velocity v is 0. Substituting the given values, we can find the maximum height and the time it takes to reach that height. This information will help us deduce a constraint on the time t at which the balls meet, which is crucial for finding t1. The steps involve algebraic manipulation, substitution, and a clear understanding of the physics of the situation.

Solving for t1: Detailed Steps

Let's delve into the detailed steps required to solve for t1. We start with the equation obtained by equating the displacements of the two balls:

44t - 4.9t² = 44(t - t1) - 4.9(t - t1)²

Expanding the terms, we get:

44t - 4.9t² = 44t - 44t1 - 4.9(t² - 2tt1 + t1²)

Further expanding and simplifying:

44t - 4.9t² = 44t - 44t1 - 4.9t² + 9.8tt1 - 4.9t1²

Notice that the 44t and -4.9t² terms cancel out on both sides, leaving us with:

0 = -44t1 + 9.8tt1 - 4.9t1²

Divide the entire equation by t1 (assuming t1 is not zero):

0 = -44 + 9.8t - 4.9t1

Rearranging the terms, we get:

4.9t1 = 9.8t - 44

Now, we need to find a relationship for t. We know that the time it takes for the first ball to reach its maximum height is given by v = u + at, where v = 0, u = 44 m/s, and a = -9.8 m/s²:

0 = 44 - 9.8t

Solving for t, we get:

t = 44 / 9.8 ≈ 4.49 seconds

This is the time it takes for the first ball to reach its maximum height. The total time the first ball is in the air before returning to the ground is twice this value, approximately 8.98 seconds. The balls must meet before the first ball returns to the ground, so we know that the meeting time t must be less than 8.98 seconds. However, the most crucial insight here is that the balls will meet when the second ball is on its way up and the first ball is on its way down, implying that they meet at half the time it takes for the first ball to go up and come down, or simply the time it takes the first ball to reach its maximum height. This is because both balls have the same initial velocity and are under the same gravitational acceleration. Therefore, they will meet at the halfway point in terms of time. Hence, t = 4.49 seconds. Now, substitute this value of t into the equation 4.9t1 = 9.8t - 44:

4.9t1 = 9.8(4.49) - 44

4.9t1 = 44.002 - 44

4.9t1 ≈ 0.002

This is an unexpected result, and it highlights the importance of critical analysis in problem-solving. The value of t1 we obtained is extremely small, close to zero, which doesn't make physical sense given the context of the problem. This discrepancy indicates that our assumption about the meeting point being at the maximum height might be incorrect, or there may be a subtle misunderstanding in our interpretation of the problem conditions. Let's re-examine our approach. The key is to recognize that the balls meet at the same height, but not necessarily at the maximum height of the first ball's trajectory. We need to find a more robust condition for when they meet.

Going back to the equation 4.9t1 = 9.8t - 44, we can express t in terms of t1:

t = (4.9t1 + 44) / 9.8

Now, we need another independent equation to solve for t1. We can use the fact that the average velocity of the first ball up to the meeting point is equal to the displacement divided by time, and similarly for the second ball. However, this approach will still lead to complex algebraic manipulations. A simpler approach is to consider the relative motion between the two balls. The second ball is thrown t1 seconds later, and they meet at some point. This means that the time for which the second ball travels before they meet is t - t1. The key insight is that when they meet, the vertical displacement is the same for both balls. Let's reconsider the equations for displacement:

s = 44t - 4.9t² (for the first ball)

s = 44(t - t1) - 4.9(t - t1)² (for the second ball)

Equating the two, we already simplified to get: 4.9t1 = 9.8t - 44. The mistake we made earlier was assuming a specific value for t. Instead, we need to find t1 such that the two balls meet. The physical condition that we haven't fully utilized is that the meeting point must be below the maximum height reached by the first ball. The maximum height can be calculated using v² = u² + 2as with v = 0:

0 = 44² - 2(9.8)s

s = 44² / (2 * 9.8) ≈ 98.78 meters

So, the meeting point must be below 98.78 meters. However, this condition alone does not help us solve for t1 directly. Let's think about the relative motion again. Consider the motion of the second ball relative to the first. The initial relative velocity is 0, as both balls are thrown with the same initial velocity. The relative acceleration is also 0, as both are under the same gravitational acceleration. This means that the relative velocity between the two balls remains constant at 0! This is a crucial insight. If the relative velocity is 0, it means that the two balls are always moving at the same velocity relative to the ground. The only way this can happen is if the time t is exactly twice the time t1. In other words, the time at which they meet is twice the time delay between the throws.

t = 2t1

Now we have two equations:

  1. 4. 9t1 = 9.8t - 44
  2. t = 2t1

Substitute the second equation into the first:

3. 9t1 = 9.8(2t1) - 44

4. 9t1 = 19.6t1 - 44

5. 7t1 = 44

t1 = 44 / 9.8 ≈ 4.49 seconds

This result is more reasonable. The time delay between the throws is approximately 4.49 seconds. This also means that the time at which they meet is t = 2t1 ≈ 8.98 seconds. This makes physical sense because it's close to the total time the first ball spends in the air.

Determining the Relative Velocity

Now that we have found t1, let's determine the relative velocity of the second ball with respect to the first ball when they pass each other. As we deduced earlier, the relative velocity between the two balls remains constant at 0 m/s throughout their motion. This is because both balls experience the same gravitational acceleration, and their initial velocities are identical. Therefore, the velocity difference between the two balls remains zero at all times, including the moment they pass each other. This elegant result simplifies the problem significantly. We don't need to calculate individual velocities at the meeting point; the relative velocity is simply 0 m/s.

Summary and Conclusion

In this article, we have successfully analyzed the motion of two balls thrown vertically into the air with an initial velocity of 44 m/s, with the second ball thrown t1 seconds after the first. Through careful application of kinematic equations and a critical examination of the problem's physical constraints, we determined that the time delay t1 is approximately 4.49 seconds. We also found that the relative velocity of the second ball with respect to the first ball when they pass each other is 0 m/s. This problem highlights the importance of a solid understanding of projectile motion principles, including constant acceleration, relative motion, and the careful interpretation of physical conditions. The initial attempt to solve for t1 led to an unrealistic result, which emphasized the need for critical thinking and re-evaluation of assumptions in problem-solving. The key insight in solving this problem was recognizing that the relative velocity between the two balls remains zero due to the identical initial velocities and accelerations. This allowed us to establish a relationship between the time delay t1 and the meeting time t, ultimately leading to the correct solution. Understanding these concepts is crucial not only for academic problem-solving but also for real-world applications involving projectile motion, such as sports, engineering, and physics research. This exploration not only provides a quantitative solution but also reinforces the analytical skills necessary for tackling complex physics problems.

Projectile motion, kinematic equations, relative velocity, time delay, acceleration due to gravity, vertical motion, displacement, initial velocity, meeting point, physics problem-solving