Question

Two cannonballs are shot from different cannons at angles \(\theta_{01}=20^{\circ}\) and \(\theta_{02}=30^{\circ}\), respectively. Assuming ideal projectile motion, the ratio of the launching speeds, \(v_{01} / v_{02},\) for which the two cannonballs achieve the same range is a) \(0.742 \mathrm{~m}\) d) \(1.093 \mathrm{~m}\) b) \(0.862 \mathrm{~m}\) e) \(2.222 \mathrm{~m}\) c) \(1.212 \mathrm{~m}\)

Short answer

Answer: (b) 0.862

Step-by-step solution

Write down the range of projectile formula

In general, the range R of an object in projectile motion can be represented by the following equation: \(R = \frac{v^2 \sin 2\theta}{g}\) where \(v\) is the initial velocity, \(\theta\) is the launch angle, and \(g\) is the acceleration due to gravity.

Define ranges for the two cannonballs

Let's represent the range of the first cannonball \(R_1\) and the range of the second cannonball \(R_2\). According to the problem, \(R_1 = R_2\).

Write down the ranges for the two cannonballs

Using the formula from step 1, let's write down the expressions for the ranges of the two cannonballs: \(R_1 = \frac{v_{01}^2 \sin 2\theta_{01}}{g}\) \(R_2 = \frac{v_{02}^2 \sin 2\theta_{02}}{g}\) Replace the angles with the given angles: \(R_1 = \frac{v_{01}^2 \sin 2 \cdot 20^\circ}{g}\) \(R_2 = \frac{v_{02}^2 \sin 2 \cdot 30^\circ}{g}\)

Set the ranges equal and solve for the ratio

Now, since \(R_1 = R_2\), let's set the two ranges equal and solve for the ratio \(\frac{v_{01}}{v_{02}}\): \(\frac{v_{01}^2 \sin 2 \cdot 20^\circ}{g} = \frac{v_{02}^2 \sin 2 \cdot 30^\circ}{g}\) Divide by g on both sides and then, divide both sides of the equation by \(v_{02}^2\) and by \(\sin2\cdot30^\circ\): \(\frac{v_{01}^2}{v_{02}^2} = \frac{\sin2\cdot20^\circ}{\sin2\cdot30^\circ}\) Taking the square root of both sides, we get: \(\frac{v_{01}}{v_{02}} = \sqrt{\frac{\sin2\cdot20^\circ}{\sin2\cdot30^\circ}}\)

Calculate the final answer and match with choices

Now, plug in the angles and find the final answer: \(\frac{v_{01}}{v_{02}} = \sqrt{\frac{\sin40^\circ}{\sin60^\circ}} \approx 0.862\) Matching our result with the given choices, the correct answer is (b) \(0.862\).

Key concepts

Range formula

In the world of physics, projectile motion is fundamentally characterized by the range formula, which is crucial for determining how far an object will travel once launched. The range formula for any object projected at an angle \[ R = \frac{v^2 \sin 2\theta}{g} \]Here:

• \( R \) is the range of the projectile, or the horizontal distance traveled.
• \( v \) is the initial velocity or speed at which the projectile is launched.
• \( \theta \) is the launch angle, the angle between the direction of launch and the horizontal plane.
• \( g \) represents the acceleration due to gravity, typically approximated as \( 9.81 \, \text{m/s}^2 \) on the surface of Earth.

Understanding this formula helps us to predict the behavior of a projectile under different conditions of launch speed and angle. It is important to note how the formula combines trigonometry and physics principles, using the function \( \sin 2\theta \) to account for directional influence. By comprehending this formula, we can derive deeper insights into how launch variables will affect the projectile's path.

Launch angle

The launch angle, symbolized by \( \theta \) in physics, significantly affects the trajectory and range of a projectile. The relationship between the launch angle and range is uniquely given through the sine of double the angle, as seen in the range formula \( \sin(2\theta) \). This relationship highlights that:

• Different launch angles can result in different ranges. The ideal launch angle in a vacuum (no air resistance) to achieve the maximum range is \( 45^{\circ} \).
• Angles less than \( 45^{\circ} \) offer less vertical height, potentially shortening the distance traveled.
• Angles greater than \( 45^{\circ} \) increase height but decrease horizontal velocity, again reducing the range.

In our exercise, two cannonballs were launched at different angles, \( 20^{\circ} \) and \( 30^{\circ} \), which inevitably affects the mechanics of achieving the same range. Using the formula, by equating the ranges, we were able to find the necessary velocity ratio, showing the interplay between launch angles and initial velocities to maintain original distance coverage.

Velocity ratio

In the comparative study of projectiles, the velocity ratio is critical for understanding how changes in the initial launching speed can offset changes in other variables such as the launch angle. The velocity ratio is represented by:\[ \frac{v_{01}}{v_{02}} \]Where:

• \( v_{01} \) is the initial velocity of the first projectile.
• \( v_{02} \) is the initial velocity of the second projectile.

In the context of this problem, to ensure that both projectiles achieve equal distances (or range), the calculation must incorporate adjustments in velocity as determined by their respective launch angles. By applying the range formula and setting the ranges equal, we solved:\[ \frac{v_{01}}{v_{02}} = \sqrt{\frac{\sin 40^{\circ}}{\sin 60^{\circ}}} \]Upon simplifying, this results in a velocity ratio of approximately 0.862. This tells us that the first projectile must be launched at about 86.2% of the speed of the second projectile to reach the same range, considering their different launch angles.