Question

On a battlefield, a cannon fires a cannonball up a slope, from ground level, with an initial velocity \(v_{0}\) at an angle \(\theta_{0}\) above the horizontal. The ground itself makes an angle \(\alpha\) above the horizontal \(\left(\alpha<\theta_{0}\right)\). What is the range \(R\) of the cannonball, measured along the inclined ground? Compare your result with the equation for the range on horizontal ground (equation 3.25\()\)

Short answer

Answer: When the angle of inclination is increased, the range on inclined ground is affected as it is dependent on the angle of inclination. Specifically, as the angle of inclination increases, the horizontal component of velocity along the inclined plane decreases while the vertical component of velocity increases. This leads to a shorter time of flight and a smaller horizontal displacement, ultimately resulting in a decrease in the range on inclined ground compared to the range on horizontal ground.

Step-by-step solution

Find the components of initial velocity

Initially, we need to find the components of initial velocity along and perpendicular to the inclined ground. The horizontal component of initial velocity \(v_{0x}\) is given by \(v_{0}\cos\theta_{0}\). The vertical component of initial velocity \(v_{0y}\) is given by \(v_{0}\sin\theta_{0}\).

Modify the components with respect to inclined ground

Next, we modify these components according to the slope of the ground. For the horizontal component of velocity along the inclined plane, \(v_{0x'}\), we have: \(v_{0x'}=v_{0x}\cos\alpha-v_{0y}\sin\alpha=v_{0}\left(\cos\theta_{0}\cos\alpha-\sin\theta_{0}\sin\alpha\right)\). For the vertical component of velocity perpendicular to inclined plane \(v_{0y'}\), we have: \(v_{0y'}=v_{0x}\sin\alpha-v_{0y}\cos\alpha=v_{0}\left(\cos\theta_{0}\sin\alpha-\sin\theta_{0}\cos\alpha\right)\).

Find the time of flight

Now, we find the time of flight of the projectile on the inclined ground, which depends on the vertical motion. The vertical motion equation is given by \(h=v_{0y'}t-\frac{1}{2}gt^2\), where h is the vertical displacement and g is the acceleration due to gravity. Since the ground is inclined, the vertical displacement is equal to 0. Thus, we can write: \(0=v_{0y'}t-\frac{1}{2}gt^2\) Solving for \(t\), we get: \(t=\frac{2v_{0y'}}{g}\)

Calculate the range on inclined ground

Now, we have the components of velocity along the inclined plane and the time of flight. The range \(R\) along the inclined ground can be found using the horizontal motion equation: \(R=v_{0x'}\cdot t\) Substituting the expressions we found for \(v_{0x'}\) and \(t\), we get: \(R=\left(v_{0}\left(\cos\theta_{0}\cos\alpha-\sin\theta_{0}\sin\alpha\right)\right)\left(\frac{2v_{0}\left(\cos\theta_{0}\sin\alpha+\sin\theta_{0}\cos\alpha\right)}{g}\right)\)

Compare with range on horizontal ground

The formula for the range of a projectile on horizontal ground is given by: \(R_{0}=\frac{v_{0}^2\sin2\theta_{0}}{g}\) The range on the inclined ground \(R\) and horizontal ground \(R_0\) may look different, but they are related through the angle of inclination \(\alpha\). In the case when the ground is horizontal (\(\alpha=0\)), the formula for the range on inclined ground reduces to the formula for range on horizontal ground: \(R=R_0=\frac{v_{0}^2\sin2\theta_{0}}{g}\) for \(\alpha=0\) In summary, we have found the expression for the range on the inclined ground and compared it with the range on horizontal ground. Depending on the angle of inclination, the range on inclined ground will be adjusted accordingly compared to the range on horizontal ground.

Key concepts

Initial Velocity Components

Understanding the components of initial velocity is crucial when it comes to problems involving projectile motion on an inclined plane. Let's take the scenario where a cannon fires a ball; the initial velocity, denoted as \(v_{0}\), is the speed with which the projectile is launched. This initial velocity can be decomposed into two perpendicular components: the horizontal component \(v_{0x}\) and the vertical component \(v_{0y}\).

The horizontal component is aligned with the ground and is calculated using \(v_{0}\cos\theta_{0}\), where \(\theta_{0}\) is the angle of projection above the horizontal. Conversely, the vertical component, which is perpendicular to the ground, is given by \(v_{0}\sin\theta_{0}\).

To address the slope of the ground, the components must be further modified. For motion along the inclined plane, the new horizontal component \(v_{0x'}\) takes the angle of the incline, \(\alpha\), into account. It's derived from a combination of \(v_{0x}\) and \(v_{0y}\) components and incorporates both the cosines and sines of the respective angles. The same goes for the new vertical component \(v_{0y'}\), which is essential in calculating subsequent projectile motion parameters such as time of flight and the range.

Time of Flight Calculation

Time of flight is a fundamental aspect of projectile motion, representing the total time a projectile spends in the air. Calculating it on an inclined plane requires a focus on vertical motion. Here, the modified vertical component of velocity, \(v_{0y'}\), plays a key role.

The equation for vertical motion in this context is \(h=v_{0y'}t-\frac{1}{2}gt^2\), with \(h\) representing the vertical displacement, \(t\) the time, and \(g\) the acceleration due to gravity. Since what goes up must come down, the displacement at the beginning and end is zero when the projectile lands back on the slope, simplifying the equation considerably.

From this equation, we can solve for the time of flight \(t\), which is the period from launch to impact on the slope. The solution shows that time of flight is directly proportional to the vertical component of the initial velocity and inversely proportional to the acceleration due to gravity.

Range of Projectile

The range of a projectile on an inclined plane, symbolized by \(R\), signifies the distance measured along the plane that the projectile travels before hitting the ground. To find it, one must combine the altered horizontal component of initial velocity with the time of flight.

The horizontal motion equation, \(R=v_{0x'}\cdot t\), provides the formula for computing range. The expression for the modified horizontal velocity component, \(v_{0x'}\), which has been adapted to the incline's angle, is factored into the equation along with the time of flight, allowing for the calculation of the projectile's range.

This range is contingent on factors like the initial speed, the angles of projection and inclination, and the force of gravity. The comparison between ranges on inclined and horizontal planes reveals that the incline modifies the projectile's trajectory and, therefore, its range.

Angle of Projection

The angle of projection, \(\theta_{0}\), is the angle at which a projectile is launched above the horizontal and significantly affects the path and distance a projectile travels. In the context of an inclined plane, this angle, coupled with the slope angle \(\alpha\), determines the initial velocity's components and consequently the projectile's motion.

When a projectile is fired, the choice of \(\theta_{0}\) influences both the magnitude and direction of the velocity components. An ideal angle in a vacuum for maximum range on flat ground is 45 degrees; however, on an inclined plane, this angle changes. The intricate relationship between the angle of projection and the plane's slope angle must be precisely calculated to determine the actual distance traveled along the inclined surface.

Balancing the angle of projection with the incline is essential for accurately predicting the behavior of a projectile. This interplay is crucial for applications ranging from simple physics exercises to real-world situations in engineering and artillery.