Question

Suppose you wish to fire a projectile over horizontal ground from the origin and attain a range of \(1000 \mathrm{m}\). a. Make a graph of the initial speed required for all firing angles \(0 < \alpha < \pi / 2\) b. What firing angle requires the least initial speed? c. What firing angle requires the least flight time?

Short answer

Answer: The firing angle that requires the least initial speed is 45 degrees, and the firing angle that requires the least flight time is 90 degrees.

Step-by-step solution

Part a: Graph the initial speed required for all firing angles 0 < α < π / 2

In order to derive the equation for the initial speed required to reach the range of 1000 meters, we can use the equation for the horizontal range of a projectile: \(Range = R = \frac{v^2 \sin{2\alpha}}{g}\) Where \(v\) is the initial speed, \(\alpha\) is the firing angle, and \(g\) (the acceleration due to gravity) is approximately \(9.81 \mathrm{m/s^2}\). We want to find the initial speed, so we can rearrange the equation for \(v\): \(v = \sqrt{\frac{R \cdot g}{\sin{2\alpha}}}\) Since we are given that \(R = 1000 \mathrm{m}\), we can plug this into the equation and plot the initial speed \(v\) for all firing angles \(0 < \alpha < \frac{\pi}{2}\).

Part b: Firing angle that requires the least initial speed

To find the angle that requires the least initial speed, we can take the derivative of the initial speed equation with respect to the angle and find the angle where the derivative is equal to 0 (i.e., finding the minimum value). So: \(\frac{dv}{d\alpha} = \frac{d}{d\alpha} \sqrt{\frac{1000 \cdot 9.81}{\sin{2\alpha}}}\) After taking the derivative and setting it to 0, we can solve for the angle: \(0 = \frac{d}{d\alpha} \sqrt{\frac{9810}{\sin{2\alpha}}}\) We find that: \(\alpha = \frac{\pi}{4}\) The firing angle that requires the least initial speed is \(\frac{\pi}{4}\), or 45 degrees.

Part c: Firing angle that requires the least flight time

To find the angle that requires the least flight time, we know that the flight time for a projectile is given by: Flight time = \(t = \frac{2v\sin\alpha}{g}\) We should take the derivative of flight time with respect to the angle and find the angle where the derivative is equal to 0 (finding the minimum value). So: \(\frac{dt}{d\alpha} = \frac{d}{d\alpha} \frac{2v\sin\alpha}{9.81}\) After taking the derivative and setting it to 0, we can solve for the angle: \(0 = \frac{d}{d\alpha} \frac{2v\sin\alpha}{9.81}\) We find that: \(\alpha = \frac{\pi}{2}\) The firing angle that requires the least flight time is \(\frac{\pi}{2}\), or 90 degrees.

Key concepts

Initial Speed

The initial speed of a projectile is the starting velocity required to reach a specific range or distance. In projectile motion, this initial speed is crucial because it determines how far the projectile will travel. To calculate the initial speed necessary for a given range, we use the following formula for horizontal range: \[ Range = R = \frac{v^2 \sin{2\alpha}}{g} \] Here, \(v\) represents the initial speed, \(\alpha\) is the firing angle, and \(g\) is the acceleration due to gravity, approximately \(9.81 \, \mathrm{m/s^2}\). This formula can be rearranged to solve for the initial speed required as: \[ v = \sqrt{\frac{R \cdot g}{\sin{2\alpha}}} \] By knowing the desired range, which is \(1000\, \mathrm{m}\) in our problem, this equation allows us to determine how fast the projectile must be launched for each possible firing angle \(0 < \alpha < \pi/2\). Plotting these values provides a comprehensive idea of how the firing angle affects the necessary initial speed.

Firing Angle

The firing angle, denoted by \(\alpha\), is the angle at which a projectile is launched. This angle significantly influences both the reach of the projectile and the initial speed required to achieve this reach. In projectile motion, the optimal angle for maximum range, with the least initial speed, stands at \(45^{\circ}\) or \(\frac{\pi}{4}\ \mathrm{radians}\).To minimize the initial speed, we determine this angle by differentiating the initial speed equation concerning \(\alpha\) and finding its minimum. Thus, \[ \alpha = \frac{\pi}{4} \] At this angle, the elements of horizontal and vertical velocity are balanced such that the projectile covers the maximum horizontal distance with the least effort. Understanding this concept helps in optimizing the launch conditions to achieve the best results.

Flight Time

Flight time refers to the duration from when a projectile is launched until it lands. It can be calculated using the following formula: \[ t = \frac{2v \sin\alpha}{g} \]Here, \(t\) denotes the flight time, \(v\) is the initial speed, \(\alpha\) is the firing angle, and \(g\) is \(9.81 \, \mathrm{m/s^2}\), the acceleration due to gravity. Flight time is mainly affected by the vertical component of the initial speed (\(v \sin\alpha\)).To find the firing angle that results in the least flight time, we set up the derivative of the flight time equation with respect to \(\alpha\) and find when it's zero. The calculation shows that \[ \alpha = \frac{\pi}{2} \] This implies a straight-up trajectory resulting in the shortest time aloft, as the projectile spends all its travel time combating gravity. Exploring the interplay between firing angle and flight time deepens our understanding of projectile dynamics.