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Mobile App Chapter 3: Problem 63
Prove that a projectile launched on level ground reaches maximum height midway along its trajectory,
Short Answer
Expert verified
Therefore, a projectile launched on level ground reaches its maximum height midway along its trajectory.
Step by step solution
01
Formulate the equations of motion
First, consider the motion of a projectile launched on level ground. This motion can be divided into two components: horizontal and vertical. Using the equations of motion, the horizontal distance (x) covered is given by \(x = v_{0} \cdot t \cdot cos(\theta)\) and the vertical distance (y) is given by \(y = v_{0} \cdot t \cdot sin(\theta) - 0.5 \cdot g \cdot t^{2}\), where \(v_{0}\) is the initial velocity, \(t\) is the time, \(\theta\) is the angle of projection and \(g\) is the acceleration due to gravity.
02
Find the time at maximum height
The maximum height is reached when the vertical velocity component becomes zero. According to the first equation of motion, \(v_{0}sin(\theta) - g \cdot t_{0} = 0\), where \(t_{0}\) is the time at maximum height. So, the time at maximum height is given by \(t_{0} = \frac{v_{0}sin(\theta)}{g}\), which will be used in the next step.
03
Calculate horizontal distance at maximum height
Next, the horizontal distance covered (x_{0}) when the projectile is at its maximum height can be calculated by substituting \(t_{0}\) from step 2 into the equation for x. This yields \(x_{0} = v_{0}\cdot \frac{v_{0}sin(\theta)}{g} \cdot cos(\theta) = \frac{v_{0}^{2}sin(2\theta)}{2g}\).
04
Calculate total horizontal distance of the trajectory
The total horizontal distance, or range, of the projectile can be calculated by finding the time of flight and substituting into the equation for x. Since the projectile is launched and lands at the same height, it's time of flight (T) is \(T = \frac{2v_{0}sin(\theta)}{g}\). So, the range (R) is \(R = v_{0} \cdot \frac{2v_{0}sin(\theta)}{g} \cdot cos(\theta) = \frac{v_{0}^{2}sin(2\theta)}{g}\).
05
Compare horizontal distance at maximum height with total horizontal distance
Now, comparing the formulas for \(x_{0}\) (from step 3) and R (from step 4), we see that \(x_{0} = \frac{1}{2}R\). This implies that the maximum height is reached when the projectile is halfway through its trajectory.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Equations of Motion
The physics of projectile motion is fascinating and relies heavily on the equations of motion. These are mathematical formulas that describe the path of a moving object with an initial velocity and subject to forces such as gravity. In essence, these equations allow us to predict where and when a projectile will land, as well as its maximum height.
For example, the horizontal motion of a projectile can be expressed as:
\(x = v_{0} \cdot t \cdot \cos(\theta)\), where \(v_{0}\) is the initial velocity, \(t\) the time elapsed, and \(\theta\) the launch angle. This equation shows that the horizontal distance traveled depends on the speed, the time flying, and the angle of projection.
Vertically, the object's motion is affected by gravity, making the equation a bit more complex:
\(y = v_{0} \cdot t \cdot \sin(\theta) - 0.5 \cdot g \cdot t^{2}\). The term \(0.5 \cdot g \cdot t^{2}\) accounts for the object's acceleration due to gravity, showing that the vertical position changes with time, not at a constant rate, but increasingly faster as it falls back to Earth. Understanding these equations is a cornerstone of any analysis involving projectile motion.
For example, the horizontal motion of a projectile can be expressed as:
\(x = v_{0} \cdot t \cdot \cos(\theta)\), where \(v_{0}\) is the initial velocity, \(t\) the time elapsed, and \(\theta\) the launch angle. This equation shows that the horizontal distance traveled depends on the speed, the time flying, and the angle of projection.
Vertically, the object's motion is affected by gravity, making the equation a bit more complex:
\(y = v_{0} \cdot t \cdot \sin(\theta) - 0.5 \cdot g \cdot t^{2}\). The term \(0.5 \cdot g \cdot t^{2}\) accounts for the object's acceleration due to gravity, showing that the vertical position changes with time, not at a constant rate, but increasingly faster as it falls back to Earth. Understanding these equations is a cornerstone of any analysis involving projectile motion.
Trajectory of a Projectile
When studying how projectiles move, we often refer to their trajectory. This is the path that a projectile follows through space as a function of time. In the context of physics, a projectile's trajectory is parabolic due to the influence of gravity on its vertical motion. Without air resistance, the trajectory's shape is determined purely by inertia and gravity.
A projectile launched at an angle \(\theta\) follows this path, rising to a peak height before descending back to the ground. Initially, gravity reduces the vertical component of the projectile's velocity, which is why the object slows down as it reaches its maximum height. After this peak, the projectile accelerates towards the ground, influenced by gravity, completing the second half of its parabolic flight path. Interestingly, at any given horizontal velocity, the maximum distance – or range – of the projectile is achieved when it is launched at an angle of 45 degrees.
A projectile launched at an angle \(\theta\) follows this path, rising to a peak height before descending back to the ground. Initially, gravity reduces the vertical component of the projectile's velocity, which is why the object slows down as it reaches its maximum height. After this peak, the projectile accelerates towards the ground, influenced by gravity, completing the second half of its parabolic flight path. Interestingly, at any given horizontal velocity, the maximum distance – or range – of the projectile is achieved when it is launched at an angle of 45 degrees.
Maximum Height of a Projectile
One of the most intriguing aspects of projectile motion is how high the projectile will reach, known as its maximum height. The maximum height is attained when the vertical component of the velocity becomes zero; at that exact moment, all the kinetic energy in the vertical direction has been converted into potential energy.
Mathematically, the time at which this occurs can be calculated by setting the vertical velocity to zero and solving for time:
\(t_{0} = \frac{v_{0}\sin(\theta)}{g}\). With this time, we can plug into our vertical motion equation to find the maximum height reached.
It's important to note that the maximum height does not depend on the horizontal component of the velocity, meaning that for a given initial speed and launch angle, the maximum height will be the same regardless of the horizontal direction. Also, as shown in the textbook exercise, the projectile reaches this maximum height halfway through its flight, which is a useful piece of information for anyone analyzing projectile motion.
Mathematically, the time at which this occurs can be calculated by setting the vertical velocity to zero and solving for time:
\(t_{0} = \frac{v_{0}\sin(\theta)}{g}\). With this time, we can plug into our vertical motion equation to find the maximum height reached.
It's important to note that the maximum height does not depend on the horizontal component of the velocity, meaning that for a given initial speed and launch angle, the maximum height will be the same regardless of the horizontal direction. Also, as shown in the textbook exercise, the projectile reaches this maximum height halfway through its flight, which is a useful piece of information for anyone analyzing projectile motion.
