Understanding Vertical Projectile Motion: A Physics Example

When you toss a ball straight up into the air, you're witnessing a classic example of vertical projectile motion. This concept is fundamental in physics, helping us understand how objects move under the influence of gravity. We'll break down a common scenario: a ball thrown vertically upward with an initial velocity of 25 m/s. Let's explore its motion at different time intervals, calculating its speed and height. This exploration will solidify your understanding of key kinematic equations and the constant acceleration due to gravity.

The Physics at Play: Gravity and Acceleration

The core principle governing the motion of our tossed ball is gravity. We know that gravity is a force that pulls everything towards the center of the Earth. For objects near the Earth's surface, this pull results in a nearly constant downward acceleration. In physics, we denote this acceleration due to gravity as 'g'. Its standard value is approximately 9.8 m/s², directed downwards. This means that for every second an object is in motion, its downward velocity increases by 9.8 m/s, or its upward velocity decreases by 9.8 m/s. In our problem, the ball is thrown upward, so gravity acts against its initial motion, causing it to slow down. As the ball reaches its peak and starts to fall back down, gravity then acts with its motion, causing it to speed up in the downward direction. Understanding this interplay between initial velocity and the constant acceleration of gravity is crucial for solving any projectile motion problem. We'll be using the standard kinematic equations, which are derived from the principles of constant acceleration, to analyze the ball's journey. These equations allow us to relate displacement, initial velocity, final velocity, acceleration, and time. The direction of motion is important; we typically define upward as positive and downward as negative. Therefore, the initial velocity of 25 m/s is positive, and the acceleration due to gravity, 'g', will be taken as -9.8 m/s².

a) How fast was it moving after 2 seconds?

To determine the ball's velocity after 2 seconds, we need to consider its initial upward velocity and the effect of gravity's downward acceleration over that time. The fundamental kinematic equation that relates final velocity ($v_f$), initial velocity ($v_i$), acceleration ($a$), and time ($t$) is: $v_f = v_i + at$. In this scenario, our initial velocity ($v_i$) is 25 m/s (positive because it's upward). The acceleration ($a$) is due to gravity, which is -9.8 m/s² (negative because it acts downward, opposing the initial upward motion). The time ($t$) we are interested in is 2 seconds. Plugging these values into the equation, we get: $v_f = 25 ext{ m/s} + (-9.8 ext{ m/s}^2)(2 ext{ s})$. Calculating this gives us $v_f = 25 ext{ m/s} - 19.6 ext{ m/s}$. Therefore, after 2 seconds, the ball's velocity is 5.4 m/s. It's still moving upward, but its speed has significantly decreased due to gravity's constant pull. This positive velocity confirms that the ball is still on its way up at this point in its trajectory. The calculation highlights how gravity gradually saps the upward momentum of the ball, a crucial aspect of understanding projectile motion. This velocity is not the speed, which is the magnitude of velocity, but the velocity itself, indicating direction. The speed at this moment would be the absolute value of this velocity, which is also 5.4 m/s.

b) How high does the ball rise after 2 seconds?

Now, let's figure out the height the ball has reached after 2 seconds. To do this, we'll use another key kinematic equation that relates displacement (height, in this case), initial velocity, acceleration, and time: d = v_i t + rac{1}{2}at^2. Here, $v_i$ is the initial velocity (25 m/s), $t$ is the time (2 seconds), and $a$ is the acceleration due to gravity (-9.8 m/s²). Let's substitute these values into the equation: d = (25 ext{ m/s})(2 ext{ s}) + rac{1}{2}(-9.8 ext{ m/s}^2)(2 ext{ s})^2. First, calculate the term involving initial velocity: $(25 ext{ m/s})(2 ext{ s}) = 50 ext{ meters}$. Next, calculate the term involving acceleration: $(2 ext{ s})^2 = 4 ext{ s}^2$, and then rac{1}{2}(-9.8 ext{ m/s}^2)(4 ext{ s}^2) = rac{1}{2}(-39.2 ext{ meters}) = -19.6 ext{ meters}. Now, add the two parts together: $d = 50 ext{ meters} - 19.6 ext{ meters}$. This gives us a displacement (height) of 30.4 meters after 2 seconds. This positive value indicates that the ball is indeed 30.4 meters above its starting point and is still ascending. This calculation demonstrates how the initial upward push, combined with the gradual deceleration caused by gravity, determines the height achieved at any given moment. It's a beautiful illustration of how energy is converted and how motion is continuously influenced by external forces. The height is measured from the point of release, assuming that point is our reference for zero displacement.

c) How fast will it be after 6 seconds?

We're going to use the same fundamental equation as in part (a) to find the velocity after 6 seconds: $v_f = v_i + at$. Our initial velocity ($v_i$) remains 25 m/s, and the acceleration ($a$) is still -9.8 m/s². However, this time, the time ($t$) is 6 seconds. Let's plug these values in: $v_f = 25 ext{ m/s} + (-9.8 ext{ m/s}^2)(6 ext{ s})$. First, calculate the change in velocity due to gravity: $(-9.8 ext{ m/s}^2)(6 ext{ s}) = -58.8 ext{ m/s}$. Now, add this to the initial velocity: $v_f = 25 ext{ m/s} - 58.8 ext{ m/s}$. This results in a final velocity ($v_f$) of -33.8 m/s. The negative sign here is very important; it indicates that after 6 seconds, the ball is moving downward with a speed of 33.8 m/s. This means the ball has already reached its peak height and is now falling back towards the ground. The magnitude of the velocity (speed) is 33.8 m/s, but the velocity itself is negative, signifying its downward direction. This illustrates a crucial point in projectile motion: the object will pass its initial height on the way down, and at any point below its peak, its velocity will be negative if upward is defined as positive. The velocity at 6 seconds is significantly larger in magnitude than at 2 seconds, showing the accelerating effect of gravity as the ball descends. It's fascinating how quickly gravity can reverse the initial upward momentum.

d) How high above the ground is it at 6 seconds?

To find the ball's position after 6 seconds, we use the same displacement equation as in part (b): d = v_i t + rac{1}{2}at^2. Remember, $v_i = 25 ext{ m/s}$, $t = 6 ext{ s}$, and $a = -9.8 ext{ m/s}^2$. Let's substitute these values: d = (25 ext{ m/s})(6 ext{ s}) + rac{1}{2}(-9.8 ext{ m/s}^2)(6 ext{ s})^2. First, calculate the displacement due to the initial velocity: $(25 ext{ m/s})(6 ext{ s}) = 150 ext{ meters}$. Next, calculate the displacement due to acceleration: $(6 ext{ s})^2 = 36 ext{ s}^2$, and then rac{1}{2}(-9.8 ext{ m/s}^2)(36 ext{ s}^2) = rac{1}{2}(-352.8 ext{ meters}) = -176.4 ext{ meters}. Now, combine these two parts: $d = 150 ext{ meters} - 176.4 ext{ meters}$. This gives us a final displacement ($d$) of -26.4 meters. The negative sign indicates that after 6 seconds, the ball is 26.4 meters below its starting point. This result makes sense given our calculation in part (c), where we found that the ball was moving downward with a significant speed. It means the ball has not only fallen back to its initial height but has continued to descend past it. This is a complete picture of the ball's vertical journey under gravity. It goes up, reaches a peak, and then falls back down, accelerating all the way. The fact that it's 26.4 meters below the release point at 6 seconds shows the significant impact of gravity over time. This demonstrates the power of kinematic equations to predict the position of objects in motion, even when forces like gravity are continuously acting upon them.

Conclusion: Mastering Vertical Motion

Understanding vertical projectile motion is a cornerstone of introductory physics. By applying the basic kinematic equations and consistently using the value of gravitational acceleration (approximately -9.8 m/s² when upward is positive), we can accurately predict the velocity and position of an object at any point in its trajectory. We've seen how an initial upward velocity is gradually counteracted by gravity, causing the object to slow down, reach a peak, and then accelerate downwards. The calculations for 2 seconds and 6 seconds clearly illustrate this phenomenon, showing the ball both ascending and descending. This problem is a practical application of Newton's laws of motion and a great way to build intuition about how forces affect movement. Keep practicing these concepts, and you'll find that physics problems related to motion become much more manageable and even intuitive.

For further exploration into the fascinating world of physics and projectile motion, you can visit NASA's Science website for real-world applications and detailed explanations. You might also find the resources on The Physics Classroom incredibly helpful for reinforcing these concepts with more examples and interactive tools.