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

Projectile motion on an inclined plane is one variation of projectile motion and the trajectory can be used to determine the position of the landing object on the included plane. We will discuss;

  • path of the trajectory of the projectile
  • range of the projectile
  • maximum range
  • initial direction is perpendicular to the direction at which the projectile hits the inclided plane when its maximum range is achieved.

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 \( \displaystyle \theta = \frac{\alpha}{2} + \frac{\pi}{4} \).

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

Question

Show that the initial direction is perpendicular to the direction at 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 at which the projectile hits the included plane.

Frequently Aksed Questions

What are the formula for 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 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 to 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 that the fired projectile lands to 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?

By merging two equations, horizontal and vertical motions, the trajectory is made. 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 equations of the projectile motion. Substitute zero into y (vertical motion), then non-zero x (horizontal motion) is the maximum range.

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