Quick Guide to Max Range in Inclined Plane Motion

The Trajectory of Projectile Motion on an Inclined Plane with Maximum Range Explained

Projectile motion on an inclined plane is a fascinating variation of the classic projectile motion problem. In this scenario, the trajectory of the projectile is influenced by the angle of the inclined plane, adding an extra layer of complexity to the problem. By understanding the key concepts and equations governing this motion, we can determine the position where the projectile lands on the inclined plane and optimise the range of the projectile.

Path of the Trajectory

The path of the projectile on an inclined plane follows a parabolic trajectory, just like in the case of projectile motion on a flat surface. However, the presence of the inclined plane alters the shape of the parabola.

The projectile’s motion can be decomposed into two components: the motion along the inclined plane and the motion perpendicular to the plane. To determine the trajectory, we need to consider the initial velocity of the projectile, the angle at which it is launched, and the angle of the inclined plane.

By applying the equations of motion and trigonometry, we can calculate the position of the projectile at any given time and plot its path on the inclined plane.

Range of the Projectile

The range of the projectile is the horizontal distance it travels along the inclined plane before landing. To calculate the range, we need to determine the time of flight of the projectile and the horizontal component of its velocity.

Moreover, the time of flight can be calculated using the vertical component of the projectile’s velocity and the acceleration due to gravity. Once we have the time of flight, we can use the horizontal component of the velocity to calculate the range.

It’s important to note that the range of the projectile on an inclined plane is affected by the angle of the plane. As the angle of the plane increases, the range of the projectile decreases, assuming all other factors remain constant.

Maximum Range

Achieving the maximum range is a key objective in many projectile motion problems. In the case of an inclined plane, the maximum range occurs when the projectile is launched at a specific angle relative to the plane.

To determine the angle for maximum range, we need to consider the relationship between the initial velocity, the angle of the inclined plane, and the angle of the projectile’s launch. By differentiating the range equation with respect to the launch angle and setting it equal to zero, we can find the optimal angle that maximises the range.

Perpendicular Direction at Maximum Range

An interesting property of the maximum range on an inclined plane is that the initial direction of the projectile is perpendicular to the direction at which it hits the plane.

In other words, when the projectile achieves its maximum range, the angle between its initial velocity vector and the vector representing its velocity at the point of impact is 90 degrees. Furthermore, this perpendicular relationship can be explained by the symmetry of the projectile’s trajectory.

At the maximum range, the projectile spends an equal amount of time ascending and descending, and the velocity vectors at the launch and landing points are mirror images of each other with respect to the perpendicular line.

Understanding projectile motion on an inclined plane is crucial for solving a wide range of problems in physics and engineering. By analysing the path of the trajectory, calculating the range, and determining the conditions for maximum range, we can gain valuable insights into the behaviour of projectiles in this scenario.

Remember that the key to success in solving these problems lies in carefully applying the relevant equations, considering the initial conditions, and understanding the relationship between the angles involved. With practice and perseverance, you’ll be able to master the concepts of projectile motion on an inclined plane and tackle even the most challenging problems with confidence.

OK, let’s take a look at the following diagram.

The diagram illustrates an inclined plane that makes an angle of \( \alpha \) radians with the horizontal. A projectile is fired from \( O \), at the bottom of the incline, with a speed of \( V \ \text{m/s}\) at an angle of elevation \( \theta \) to the horizontal.

Part 1: Horizontal position of the projectile

Resolving the initial horizontal velocity, \( \dot{x} = V \cos \alpha \).

\( \begin{align} \ddot{x} &= 0 \\ \dot{x} &= C_1 &\text{integrating} \\ \dot{x} &= V \cos \theta &\text{ when } t=0 \\ V \cos \theta &= C_1 \\ \therefore \dot{x} &= V \cos \theta \\ x &= V t \cos \theta + C_2 \\ x &= 0 \text{ when } t=0 \\ 0 &= C_2 \\ \therefore x &= V t \cos \theta \cdots (1) \end{align} \)

Part 2: Vertical position of the projectile

Resolving the initial vertical velocity, \( \dot{y} = V \sin \alpha \).

\( \begin{align} \ddot{y} &= -g \\ \dot{y} &= -gt + C_3 &\text{integrating} \\ \dot{y} &= V \sin \theta &\text{ when } t = 0 \\ V \sin \theta &= C_3 \\ \therefore \dot{y} &= -gt + V \sin \theta \\ y &= \displaystyle -\frac{1}{2}gt^2 + V t \sin \theta + C_4 &\text{integrating} \\ y &= 0 &\text{ when } t = 0 \\ 0 &= C_4 \\ \therefore y &= -\frac{1}{2} gt^2 + Vt \sin \theta \cdots (2) \end{align} \)

where \( t \) is the time in seconds, after firing, and \( g \) is the acceleration due to gravity.

Part 3: Path of the trajectory of the projectile

\( \begin{align} \displaystyle t &= \frac{x}{V \cos \alpha} &\text{from } (1) \\ y &= -\frac{gx^2}{2V^2 \cos^2 \theta} + \frac{V x \sin \theta}{V \cos \theta} &\text{substituting into } (2) \\ &= -\frac{gx^2}{2V^2} \sec^2 \theta + x \tan \theta \\ \therefore y &= x \tan \theta-Kx^2 \sec^2 \theta \cdots (3) &\text{by letting } \displaystyle \frac{g}{2V^2} = K \text{ for simplicity} \end{align} \)

Part 4: Range of the projectile

\( \begin{align} y &= x \tan \alpha &\text{equation of the inclined plane} \cdots (4) \\ x \tan \alpha &= x \tan \theta-Kx^2 \sec^2 \theta &\text{equating (3) and (4)} \\ \tan \alpha &= \tan \theta-Kx \sec^2 \theta \\ Kx \sec^2 \theta &= \tan \theta-\tan \alpha \\ &= \displaystyle \frac{\sin \theta}{\cos \theta}-\frac{\sin \alpha}{\cos \alpha} \\ &= \frac{\sin \theta \cos \alpha-\cos \theta \sin \alpha}{\cos \theta \cos \alpha} \\ &= \frac{\sin (\theta-\alpha)}{\cos \theta \cos \alpha} \\ \therefore x &= \frac{\sin(\theta-\alpha)}{\cos \theta \cos \alpha} \times \frac{1}{K \sec^2 \theta} \\ &= \frac{\sin (\theta-\alpha)}{\cos \theta \cos \alpha} \times \frac{\cos^2 \theta}{K} \\ &= \frac{\sin (\theta-\alpha) \cos \theta}{K \cos \alpha} \cdots (5) \\ \cos \alpha &= \frac{x}{r} &\text{inclined plane} \\ r &= \frac{x}{\cos \alpha} \\ \therefore r &= \frac{\sin(\theta-\alpha) \cos \theta}{K \cos \alpha} \times \frac{1}{\cos \alpha} \\ &= \frac{\sin (\theta-\alpha) \cos \theta}{K \cos^2 \alpha} \cdots (6) \end{align} \)

Part 5: Maximum range, \( R \)

\( \begin{align} \displaystyle \frac{dr}{d \theta} &= \frac{\sin (\theta-\alpha) \times (-\sin \theta) + \cos \theta \times \cos (\theta-\alpha)}{K \cos^2 \alpha} &\text{product rule as } K \cos^2 \alpha \text{ is constant} \\ &= \frac{\cos \theta \cos (\theta-\alpha)-\sin \theta \sin (\theta-\alpha)}{K \cos^2 \alpha} \\ &= \frac{\cos (\theta + \theta-\alpha)}{K \cos^2 \alpha} \\ &= \frac{\cos (2 \theta-\alpha)}{K \cos^2 \alpha} \\ \frac{dr}{d \theta} &= 0 &\text{stationary point} \\ \frac{\cos (2 \theta-\alpha)}{K \cos^2 \alpha} &= 0 \\ \cos (2 \theta-\alpha) &= 0 \\ 2 \theta-\alpha &= \frac{\pi}{2} \\ \theta &= \frac{\alpha}{2} + \frac{\pi}{4} \cdots (7) \\ \frac{d^2r}{d \theta^2} &= \frac{-2 \sin (2 \theta-\alpha)}{K \cos^2 \alpha} < 0 &\text{when } 2 \theta-\alpha = \frac{\pi}{2} \end{align} \)

Therefore \( r \) has a maximum value, \( R \), when \( \displaystyle \theta = \frac{\alpha}{2} + \frac{\pi}{4} \).

\( \begin{align} \displaystyle R &= \frac{\sin(\frac{\alpha}{2} + \frac{\pi}{4}-\alpha) \cos(\frac{\alpha}{2} + \frac{\pi}{4})}{K \cos^2 \alpha} &\text{substituting into } (6) \\ &= \frac{ \sin(\frac{\pi}{4}-\frac{\alpha}{2}) \cos(\frac{\pi}{4} + \frac{\alpha}{2})}{K(1-\sin^2 \alpha)} \\ &= \frac{ 2\sin(\frac{\pi}{4}-\frac{\alpha}{2}) \cos(\frac{\pi}{4} + \frac{\alpha}{2})}{2K(1-\sin^2 \alpha)} \\ &= \frac{\sin\left[ (\frac{\pi}{4}-\frac{\alpha}{2}) + ( \frac{\pi}{4} + \frac{\alpha}{2} ) \right] + \sin \left[ ( \frac{\pi}{4}-\frac{\alpha}{2} )-( \frac{\pi}{4} + \frac{\alpha}{2} ) \right]}{2K(1 + \sin \alpha)(1-\sin \alpha)} &2 \sin x \cos y = \sin (x+y) + \sin (x-y) \\ &= \frac{\sin \frac{\pi}{2} + \sin (-\alpha)}{ 2K(1 + \sin \alpha)(1-\sin \alpha)} \\ &= \frac{1-\sin \alpha}{ 2K(1 + \sin \alpha)(1-\sin \alpha)} \\ &= \frac{1}{2K(1 + \sin \alpha)} \end{align} \)


Show that the initial direction is perpendicular to the direction in which the projectile hits the inclined plane.

\( \begin{align} \displaystyle m_1 &= \tan \theta &\text{the slope of the initial direction} \\ &= \tan \left(\frac{\alpha}{2} + \frac{\pi}{4} \right) &\text{from } (7) \\ &= \frac{\tan \frac{\alpha}{2} + \tan \frac{\pi}{4}}{1-\tan \frac{\alpha}{2} \tan \frac{\pi}{4}} \\ &= \frac{1 + \tan \frac{\alpha}{2}}{1-\tan\frac{\alpha}{2}} \cdots (8) \\ y &= x \tan \theta-K x^2 \sec^2 \theta &\text{from (3)} \\ \frac{dy}{dx} &= \tan \theta-2Kx \sec^2 \theta &\text{the slope of the direction at impact} \\ m_2 &= \tan \theta-2K \times \frac{\sin (\theta-\alpha) \cos \theta}{K \cos \alpha} \times \frac{1}{\cos^2 \theta} &x = \frac{\sin(\theta-\alpha) \cos \theta}{K \cos \alpha} \text{ from } (5) \\ &= \tan \theta-\frac{2 \sin (\theta-\alpha)}{\cos \alpha \cos \theta} \\ &= \tan \theta-\frac{2 \sin \theta \cos \alpha-2 \cos \theta \sin \alpha}{\cos \alpha \cos \theta} \\ &= \tan \theta-\frac{2 \sin \theta}{\cos \theta} + \frac{2 \sin \alpha}{\cos \alpha} \\ &= \tan \theta-2 \tan \theta + 2 \tan \alpha \\ &= 2 \tan \alpha-\tan \theta \\ &= \frac{4 \tan \frac{\alpha}{2}}{1-\tan^2 \frac{\alpha}{2}}-\frac{1 + \tan \frac{\alpha}{2}}{ 1-\tan \frac{\alpha}{2}} &\text{from (8)} \\ &= \frac{4 \tan \frac{\alpha}{2}}{1-\tan^2 \frac{\alpha}{2}}-\frac{\left(1 + \tan \frac{\alpha}{2} \right)^2}{ 1-\tan^2 \frac{\alpha}{2} } \\ &= \frac{4 \tan \frac{\alpha}{2}-\left( 1 + 2 \tan \frac{\alpha}{2} + \tan^2 \frac{\alpha}{2} \right)}{1-\tan^2 \frac{\alpha}{2}} \\ &= \frac{-1 + 2 \tan \frac{\alpha}{2}-\tan^2 \frac{\alpha}{2}}{1-\tan^2 \frac{\alpha}{2}} \\ &= -\frac{1-2 \tan \frac{\alpha}{2} + \tan^2 \frac{\alpha}{2}}{1-\tan^2 \frac{\alpha}{2}} \\ &= -\frac{\left(1-\tan \frac{\alpha}{2}\right)^2}{1-\tan^2 \frac{\alpha}{2}} \\ &= -\frac{\left(1-\tan^2 \frac{\alpha}{2}\right)^2}{\left(1-\tan \frac{\alpha}{2} \right) \left( 1 + \tan \frac{\alpha}{2} \right)} \\ &= -\frac{1-\tan \frac{\alpha}{2}}{1 + \tan \frac{\alpha}{2}} \\ m_1 \times m_2 &= \frac{1 + \tan \frac{\alpha}{2}}{1-\tan \frac{\alpha}{2}} \times -\frac{1-\tan \frac{\alpha}{2}}{1 + \tan \frac{\alpha}{2}} \\ &= -1 \end{align} \)

Therefore the initial direction is perpendicular to the direction in which the projectile hits the included plane.

Frequently Asked Questions

What is the formula for a projectile on an inclined plane?

The formula of a projectile on an inclined plane is similar to what’s on a plane. The equation of the inclined plane (a straight line) is used to find the position on the inclined plane by equating the inclined plane equation and the projectile motion.

How is projectile motion on a plane?

The vertical gravitational acceleration is set at its initial value, and integrating twice gives the vertical equation of the projectile. The horizontal acceleration is always zero, while no air resistance is involved in the horizontal motion. Integrating twice from this acceleration gives the equation of the horizontal motion of the projectile.

What is inclined projectile motion?

An inclined projectile motion is when the fired projectile lands on an inclined plane where its inclination angle is between 0 and 90 degrees. Therefore this angle of inclination makes the equations of the projectile motion.

What does trajectory mean?

The trajectory is made by merging two equations, horizontal and vertical motions. When merging them, removing the angle parameter makes one equation that consists of only x (horizontal) and y (vertical) parameters.

What is the formula for maximum range?

The maximum range occurs when the firing angle is 45 degrees. So set the firing angle to 45 degrees and substitute this angle into the projectile motion equations. Substitute zero into y (vertical motion), and then non-zero x (horizontal motion) is the maximum range.

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