Calculating Maximum Height Projectile Motion Physics Problem

Motion in physics often involves objects moving under the influence of gravity, and one classic example is a body thrown vertically upwards. Understanding the principles governing this motion allows us to determine key parameters like the maximum height reached. In this comprehensive exploration, we will dissect the physics behind projectile motion, focusing on how to calculate the maximum height attained by an object thrown upwards, given its height equation. Let's delve into the concepts and calculations involved in solving this fascinating problem.

Decoding the Height Equation

In analyzing the motion of a body thrown vertically upwards, the height, ${ h }$, attained in time, ${ t }$, is described by the equation:

${ h = 20t - \frac{49}{10} t^2 }$

This equation is a quadratic function, which makes sense because the motion is governed by constant acceleration due to gravity. The equation itself tells a story of the projectile's journey: the term ${ 20t }$ indicates an initial upward velocity, while the term ${ -\frac{49}{10} t^2 }$ represents the effect of gravity slowing the body down as it rises. The negative sign signifies that gravity acts in the opposite direction to the initial velocity, causing the body to decelerate as it moves upwards. The coefficient ${ \frac{49}{10} }$ is related to the acceleration due to gravity, which is approximately ${ 9.8 \text{ m/s}^2 }$. Let’s break down the components to understand their physical significance.

Initial Velocity and Gravity's Pull

The initial term, ${ 20t }$, is crucial as it represents the initial upward thrust given to the body. This means the body was thrown upwards with an initial velocity of 20 meters per second. Without this initial push, the body would simply remain at its starting point, influenced only by gravity. This initial velocity is what propels the body against gravity, allowing it to gain altitude. As time progresses, this term increases the height, but its effect is countered by the gravitational force pulling the body downwards. The interplay between the initial velocity and gravity is what dictates the trajectory of the body.

Conversely, the second term, ${ -\frac{49}{10} t^2 }$, is the gravitational component. This term embodies the effect of gravity on the body, pulling it back towards the Earth. The negative sign indicates that gravity acts in the opposite direction to the initial upward motion, which is why the term subtracts from the overall height. The factor ${ \frac{49}{10} }$ is derived from half of the acceleration due to gravity (${ g }$), which is approximately ${ 9.8 \text{ m/s}^2 }$. This factor means that for every second that passes, gravity reduces the height by an increasing amount, proportional to the square of the time. This quadratic relationship illustrates the non-linear effect of gravity, causing the body to slow down more and more as it ascends.

The Significance of the Quadratic Form

The quadratic form of the equation, ${ h = 20t - \frac{49}{10} t^2 }$, is particularly significant because it tells us that the height, ${ h }$, follows a parabolic path over time, ${ t }$. In mathematical terms, a parabola is a U-shaped curve, and in this physical context, it represents the trajectory of the body as it moves upwards and then falls back down. The initial upward motion, represented by the positive linear term (${ 20t }$), contributes to the upward curve of the parabola, while the gravitational effect, represented by the negative quadratic term (${ -\frac{49}{10} t^2 }$), curves the path downwards. This parabolic path is a hallmark of projectile motion under constant gravitational acceleration.

The vertex of this parabola represents the maximum height reached by the body. At this point, the body momentarily stops moving upwards before it begins to descend. Finding the vertex of the parabola is thus equivalent to finding the maximum height. The vertex is the point where the slope of the curve changes from positive (ascending) to negative (descending), indicating the transition from upward to downward motion. Understanding the properties of parabolas and quadratic equations is crucial for solving problems related to projectile motion and finding key parameters like maximum height.

Determining Maximum Height: Methods and Techniques

To determine the maximum height reached by the body, we can employ a couple of methods. Both approaches leverage the principles of calculus and the properties of quadratic functions to pinpoint the highest point in the projectile's trajectory. These methods provide us with the tools to dissect the height equation and extract valuable information about the body's motion.

Method 1: Using Calculus (Finding the Vertex)

The first method involves the use of calculus. The maximum height occurs at the vertex of the parabolic path, which corresponds to the point where the vertical velocity of the body is momentarily zero. This is the instant when the body transitions from moving upwards to moving downwards. Calculus provides us with the tools to find this critical point by determining when the derivative of the height function with respect to time is equal to zero.

To begin, we differentiate the height equation with respect to time, ${ t }$. The derivative, ${ \frac{dh}{dt} }$, gives us the vertical velocity of the body at any given time. Differentiating the equation:

${ h = 20t - \frac{49}{10} t^2 }$

we get:

${ \frac{dh}{dt} = 20 - \frac{49}{5} t }$

This equation represents the instantaneous vertical velocity of the body as a function of time. The first term, 20, is the initial upward velocity, while the second term, ${ -\frac{49}{5} t }$, represents the decrease in velocity due to gravity. Setting ${ \frac{dh}{dt} = 0 }$ allows us to find the time at which the body’s vertical velocity is zero, which is the moment it reaches its maximum height. So, we solve:

${ 20 - \frac{49}{5} t = 0 }$

Solving for ${ t }$:

${ t = \frac{20}{\frac{49}{5}} = \frac{20 \times 5}{49} = \frac{100}{49} \text{ seconds} }$

This value of ${ t }$, approximately 2.04 seconds, is the time at which the body reaches its maximum height. To find the maximum height itself, we substitute this value of ${ t }$ back into the original height equation:

${ h_{max} = 20 \left( \frac{100}{49} \right) - \frac{49}{10} \left( \frac{100}{49} \right)^2 }$

Calculating this gives us:

${ h_{max} = \frac{2000}{49} - \frac{49}{10} \times \frac{10000}{49^2} = \frac{2000}{49} - \frac{1000}{49} = \frac{1000}{49} \text{ meters} }$

Thus, the maximum height reached by the body is approximately 20.41 meters. This method utilizes the power of calculus to pinpoint the exact moment when the body's vertical velocity is zero, thereby allowing us to calculate the maximum height with precision.

Method 2: Completing the Square (Algebraic Approach)

An alternative method to find the maximum height involves an algebraic technique known as completing the square. This method transforms the quadratic equation into a form that directly reveals the vertex of the parabola, providing a clear path to determining the maximum height without the need for calculus. Completing the square is a powerful tool in algebra, allowing us to rewrite quadratic expressions in a more insightful way.

Starting with the height equation:

${ h = 20t - \frac{49}{10} t^2 }$

We rearrange the terms to have the quadratic term first:

${ h = -\frac{49}{10} t^2 + 20t }$

To complete the square, we factor out the coefficient of the ${ t^2 }$ term, which is ${ -\frac{49}{10} }$:

${ h = -\frac{49}{10} \left( t^2 - \frac{200}{49} t \right) }$

Now, we need to add and subtract a value inside the parentheses to create a perfect square trinomial. The value we need to add and subtract is half of the coefficient of ${ t }$, squared. The coefficient of ${ t }$ inside the parentheses is ${ -\frac{200}{49} }$, so half of it is ${ -\frac{100}{49} }$, and squaring this gives us ${ \left( \frac{100}{49} \right)^2 }$. Therefore, we add and subtract this inside the parentheses:

${ h = -\frac{49}{10} \left( t^2 - \frac{200}{49} t + \left( \frac{100}{49} \right)^2 - \left( \frac{100}{49} \right)^2 \right) }$

We can now rewrite the first three terms inside the parentheses as a perfect square:

${ h = -\frac{49}{10} \left( \left( t - \frac{100}{49} \right)^2 - \left( \frac{100}{49} \right)^2 \right) }$

Distribute the ${ -\frac{49}{10} }$ back into the parentheses:

${ h = -\frac{49}{10} \left( t - \frac{100}{49} \right)^2 + \frac{49}{10} \left( \frac{100}{49} \right)^2 }$

Simplify the second term:

${ h = -\frac{49}{10} \left( t - \frac{100}{49} \right)^2 + \frac{1000}{49} }$

This equation is now in the vertex form of a parabola, ${ h = a(t - k)^2 + h_v }$, where ${ (k, h_v) }$ is the vertex of the parabola. In our case, the vertex is at ${ \left( \frac{100}{49}, \frac{1000}{49} \right) }$. The maximum height, ${ h_v }$, is the y-coordinate of the vertex, which is ${ \frac{1000}{49} }$ meters, or approximately 20.41 meters. The x-coordinate, ${ \frac{100}{49} }$, is the time at which the maximum height is reached, which we also found using calculus. This algebraic method provides a clear and direct way to find the maximum height by rewriting the quadratic equation in vertex form.

Conclusion: The Apex of Projectile Motion

In summary, finding the maximum height of a body thrown vertically upwards involves understanding the interplay between initial velocity and gravitational acceleration. Whether we employ the elegance of calculus by finding where the derivative of the height function is zero or utilize the algebraic power of completing the square, the result is the same: the maximum height is reached when the body's upward motion is momentarily halted by gravity before it begins its descent.

Both methods have their merits. The calculus approach gives us a deeper insight into the dynamics of the motion, emphasizing the concept of velocity and its rate of change. By setting the derivative of the height function to zero, we directly address the condition for maximum height – the point where the vertical velocity is zero. This method is particularly useful in more complex scenarios where velocities and accelerations may vary with time or position.

On the other hand, completing the square offers a purely algebraic route to the solution, sidestepping the need for calculus. This method highlights the properties of quadratic equations and parabolas, transforming the equation into a form that directly reveals the vertex, and hence the maximum height. It's a powerful reminder of how algebraic manipulations can simplify problem-solving and provide alternative perspectives on physical phenomena.

In this specific problem, the maximum height reached by the body is ${ \frac{1000}{49} }$ meters, or approximately 20.41 meters. This height is a critical parameter in understanding the projectile's trajectory and can be used to further analyze the motion, such as calculating the total time of flight or the velocity at impact. Understanding how to calculate such parameters is essential not only in physics but also in various engineering and sports applications.

By mastering the techniques to analyze projectile motion, we gain valuable insights into the world around us, from the simple act of throwing a ball to more complex scenarios like satellite orbits and ballistic trajectories. The principles discussed here form the foundation for more advanced topics in mechanics and dynamics, making the study of projectile motion a cornerstone in physics education.