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Chapter 3: Problem 31

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

Expert verified
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

01

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 \(\theta_{0}\). \(v_{x_{0}} = v_{0} \cos{\theta_{0}}\) \(v_{y_{0}} = v_{0} \sin{\theta_{0}}\)
02

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 \(t\): \(0 = v_{y_{0}}t - \frac{1}{2}gt^2\) \(t = \frac{2v_{y_{0}}}{g}\)
03

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: \(x = v_{x_{0}}\frac{2v_{y_{0}}}{g} = \frac{2v_{0}^2 \sin{\theta_{0}} \cos{\theta_{0}}}{g}\)
04

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: \(R = \frac{2v_{0}^2 \sin{\theta_{0}} \cos{\theta_{0}}}{g\cos{\alpha}}\)
05

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.

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