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

Question: Calculate the range of a cannonball fired up a slope with an angle α, given an initial velocity v₀ and firing angle θ₀. Compare the result with the range equation for horizontal ground. Answer: The range R can be calculated using the following equation: \(R = \frac{2v_{0}^2 \sin{\theta_{0}} \cos{\theta_{0}}}{g\cos{\alpha}}\). Comparing this with the range equation for horizontal ground, \(R_{0} = \frac{v_{0}^2\sin{2\theta_{0}}}{g}\), we can see that the range equation for inclined ground includes an additional factor of \(\frac{1}{\cos{\alpha}}\) in the denominator to account for the effect of the incline on the range.

Step-by-step solution

Determine initial horizontal and vertical velocities

We can begin by finding the initial horizontal and vertical components of velocity, using the initial velocity \(v_{0}\) and the angle \(\theta_{0}\). \(v_{x_{0}} = v_{0} \cos{\theta_{0}}\) \(v_{y_{0}} = v_{0} \sin{\theta_{0}}\)

Calculate time of flight

In order to find the range, we first need to calculate the time of flight of the cannonball. To do this, we will use the vertical motion equation and set the final vertical position \(y\) equal to the initial height \(y_{0} = 0\). \(y = y_{0} + v_{y_{0}}t - \frac{1}{2}gt^2\) Since both initial and final vertical positions are the same (\(y = y_{0} = 0\)), we can rearrange the equation to solve for the time \(t\): \(0 = v_{y_{0}}t - \frac{1}{2}gt^2\) \(t = \frac{2v_{y_{0}}}{g}\)

Calculate horizontal displacement during the time of flight

Next, we will calculate the horizontal displacement \(x\) during the time of flight. We can use the horizontal motion equation for this, remembering that there is no horizontal acceleration: \(x = v_{x_{0}}t\) Substitute the time \(t\) found in step 2: \(x = v_{x_{0}}\frac{2v_{y_{0}}}{g} = \frac{2v_{0}^2 \sin{\theta_{0}} \cos{\theta_{0}}}{g}\)

Calculate the range along the inclined ground

Now we will find the range \(R\) along the inclined ground. Recall that the angle of the ground is \(\alpha\). We can use trigonometry to relate \(x\) to \(R\): \(R = \frac{x}{\cos{\alpha}}\) Substitute the horizontal displacement found in step 3: \(R = \frac{2v_{0}^2 \sin{\theta_{0}} \cos{\theta_{0}}}{g\cos{\alpha}}\)

Compare the result with the range equation on horizontal ground

The range equation for horizontal ground (equation 3.25) is: \(R_{0} = \frac{v_{0}^2\sin{2\theta_{0}}}{g}\) Comparing the range equation found in step 4 with the range equation for horizontal ground, we can see that they are similar, but the equation for the inclined ground contains an additional factor of \(\frac{1}{\cos{\alpha}}\) in the denominator. This factor takes into account the effect of the incline on the range of the cannonball.

Key concepts

Inclined Plane

Inclined planes are surfaces that are tilted at an angle, deviating from the horizontal level. These surfaces impact how objects move along them, which is essential in understanding projectile motion on slopes. When a projectile, like a cannonball, is launched up a slope, we aren't just concerned with its horizontal and vertical motions independently – we must also factor in the incline's effect on these motions, particularly on the range the projectile covers.

In our exercise, the incline is set at an angle \(\alpha\) above the horizontal. This means the entire ground moves upward at this angle, altering the projectile's trajectory compared to flat ground. As the projectile traverses this sloped surface, the effective force acting along the incline includes components of both gravitational pull and initial launch velocity. This results in a reduction of the horizontal reach since it's working against the incline as it travels.

Effectively, the inclined plane modifies the path trajectory, affecting both the speed at which the projectile covers distances and the total range. By analyzing these elements, you can better appreciate how incline angles play a crucial role in projectile physics.

Range Equation

The range equation is vital for predicting how far a projectile will travel. For a typical horizontal surface, the standard range equation applicable is: \[R_0 = \frac{v_0^2 \sin{2\theta_0}}{g}.\]This formula calculates range as a function of the initial velocity \(v_0\), launch angle \(\theta_0\), and gravitational acceleration \(g\).

When working with inclined planes, the range equation is modified to account for the incline angle \(\alpha\). We derive the inclined range equation using:\[R = \frac{2v_0^2 \sin{\theta_0} \cos{\theta_0}}{g\cos{\alpha}}.\]

Notice the additional term \(\cos{\alpha}\) in the denominator. This factor adjusts the range for the slope's angle, effectively showing how steeper slopes decrease projectile range compared to horizontal surfaces. Thus, the incline means the cannonball covers less horizontal distance than it would without the slope, despite having the same initial speed and launch angle.

By comprehending the range equation's adjustment for inclined surfaces, you can better navigate real-world applications where launching projections involve uneven terrain.

Trigonometric Functions

Trigonometry, the mathematics of angles, is at the heart of understanding projectile motion & physics. It allows us to split vectors into vertical and horizontal components using sine and cosine functions, which is necessary when tackling problems involving angles, like our cannonball on an inclined plane.

Trigonometric functions assist in decomposing initial velocities. Given an initial speed \(v_0\) and launch angle \(\theta_0\), the initial horizontal velocity is represented as:\[v_{x_0} = v_0 \cos{\theta_0}\]and the initial vertical velocity as:\[v_{y_0} = v_0 \sin{\theta_0}.\]

The \(\cos\alpha\) function plays a critical role too, particularly when transforming calculated horizontal displacement onto the inclined plane. In the range equation, \(\cos\alpha\) helps us adjust for the angle at which the inclined plane rises, articulating how components of motion behave along non-horizontal surfaces.

Understanding the interplay of trigonometric functions prepares you for solving real-world physics problems involving any angled plane or motion.