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For a given initial speed of an ideal projectile, there is (are) ______ launch angle(s) for which the range of the projectile is the same. a) only one b) two different c) more than two but a finite number of d) only one if the angle is \(45^{\circ}\) but otherwise two different e) an infinite number of

Short Answer

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launch angles.

Step by step solution

01

Understand the problem

First, we need to recall the equation for the range of a projectile launched at an angle \(\theta\) with an initial velocity \(v_0\). The motion must be analyzed in the horizontal (x) and vertical (y) directions.
02

Analyze the range formula

The range of a projectile is given by the equation: \[R = \frac{v_0^2 \sin (2\theta)}{g}\] Where \(R\) is the range, \(v_0\) is the initial velocity, \(\theta\) is the launch angle, and \(g\) is the acceleration due to gravity. Note that this equation has a sine function which is periodic and covers all values from \(-1\) to \(1\).
03

Analyze the sine function

We know that the sine function has the property: \[\sin(\alpha + n180^{\circ}) = \sin(\alpha)\] for every integer \(n\). In our case, we are interested in angles between \(0^{\circ}\) and \(180^{\circ}\), so we can focus on the property: \[\sin(\theta + 180^{\circ}) = \sin(\theta)\]
04

Relating the launch angles

Using the sine property above, we can relate two different angles (\(\theta_1\) and \(\theta_2\)) that result in the same range: \[\sin(2\theta_1) = \sin(2\theta_2)\] By applying the property mentioned earlier, we have: \[\sin(2\theta_2) = \sin(2\theta_1 + 180^{\circ})\] Which simplifies to: \[2\theta_2 = 2\theta_1 + 180^{\circ}\]
05

Conclusion

From the analysis above, we see that there are always two angles, \(\theta_1\) and \(\theta_2\), that result in the same range, provided that they are not equal to \(90^{\circ}\). Since the angles are always related by a \(180^{\circ}\) difference, and the maximum possible angle for a projectile is \(180^{\circ}\), we conclude that there are always two different angles for which the projectile can be launched to achieve the same range. Therefore, the correct answer is: b) two different

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Range of Projectile
The range of a projectile is one of the crucial aspects students studying projectile motion need to understand. Essentially, the range refers to the horizontal distance a projectile travels before returning to the same level from which it was projected.

Using the standard range formula \[R = \frac{v_0^2 \sin (2\theta)}{g}\], we can determine how far the object will go. The variable \(v_0\) represents the initial velocity of the projectile, \(\theta\) is the launch angle, \(g\) is the acceleration due to gravity, and \(\sin\) is the sine function from trigonometry. What's intriguing here is the sine function, which has properties that affect the range. For example, \(\sin(2\theta)\) will have the same value for two different angles within a 0 to 180-degree range, hence giving the same projectile range.

The formula indicates how sensitive the range is to both the speed it's thrown at and the angle of release. As such, participants in sports such as javelin, discus, or even basketball, intuitively understand the concept of the range of a projectile.
Launch Angle of Projectile
Understanding the launch angle of a projectile is tantamount to mastering the art of projectile motion. The launch angle, \(\theta\), is the angle at which the projectile is thrown relative to the horizontal.

Choosing the right launch angle is crucial, as it determines both the range and height of a projectile. It's like deciding where to aim when kicking a soccer ball to clear a defensive wall and beat the goalkeeper. If the launch angle is too steep or too shallow, the projectile might not reach the desired distance. Interestingly, there are two different angles that can give a projectile the same range - a mathematical sweet spot if you will.

The unique relationship between pairs of angles leading to identical ranges is a fundamental property in projectile motion. Hence, when planning trajectories, one must consider the launch angle's profound role in hitting the target. It's a concept that finds applications in everything from sports to space launches!
Sine Function in Physics
The sine function is more than just a topic in mathematics; it's a pivotal tool in physics, especially in studying wave phenomena and projectile motion. In our context, the sine function helps relate the launch angle to the range of a projectile.

For projectile motion, we generally focus on \(\sin(2\theta)\) as it appears in the range equation. The function oscillates between -1 and 1, giving us valuable insight into how the range varies with the launch angle. Regardless of the angle you choose, the sine function ensures that every launch angle \(\theta\) has a complementary angle (\(180^\circ - \theta\)) resulting in the same sine value, and thus, the same range.

Appreciating the sine function's role in physics means understanding that it elegantly bridges the gap between angles and distances, a bridge we frequently cross in analyzing projectile motion. It's a reminder that sometimes, the key to solving a complex physical problem lies in the simple sine curve.
Acceleration Due to Gravity
Acceleration due to gravity, denoted as \(g\), is a constant that signifies the rate at which objects accelerate towards the Earth when in free fall. Its standard value is approximately \(9.81 m/s^2\) on Earth's surface, although it can vary slightly depending on altitude and geological formations.

Gravity imparts a downward acceleration on every projectile, thus it plays a pivotal role in determining the trajectory and eventual landing spot of a thrown object. In our range formula, gravity is the great equalizer; it affects all projectiles the same, regardless of their mass. When solving problems related to the motion of a projectile, we always account for gravity because it's the force that eventually brings our projectile back down to Earth, literally.

Students of physics must remember that gravity is not only a force but an accelerator in motion equations, contributing to the parabolic path traced out by every heaved basketball or lobbed grenade. This understanding is vital, as it allows us to predict the landings of projectiles, whether in theoretical calculations or real-world applications.

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Most popular questions from this chapter

A truck travels \(3.02 \mathrm{~km}\) north and then makes a \(90.0^{\circ}\) left turn and drives another \(4.30 \mathrm{~km} .\) The whole trip takes \(5.00 \mathrm{~min}\). a) With respect to a two-dimensional coordinate system on the surface of Earth such that the \(y\) -axis points north, what is the net displacement vector of the truck for this trip? b) What is the magnitude of the average velocity for this trip?

A box containing food supplies for a refugee camp was dropped from a helicopter flying horizontally at a constant elevation of \(500 . \mathrm{m}\). If the box hit the ground at a distance of \(150 . \mathrm{m}\) horizontally from the point of its release, what was the speed of the helicopter? With what speed did the box hit the ground?

Some rental cars have a GPS unit installed, which allows the rental car company to check where you are at all times and thus also know your speed at any time. One of these rental cars is driven by an employee in the company's lot and, during the time interval from 0 to \(10.0 \mathrm{~s}\), is found to have a position vector as a function of time given by $$ \begin{aligned} \vec{r}(t)=&\left((24.4 \mathrm{~m})-t(12.3 \mathrm{~m} / \mathrm{s})+t^{2}\left(2.43 \mathrm{~m} / \mathrm{s}^{2}\right)\right.\\\ &\left.(74.4 \mathrm{~m})+t^{2}\left(1.80 \mathrm{~m} / \mathrm{s}^{2}\right)-t^{3}\left(0.130 \mathrm{~m} / \mathrm{s}^{3}\right)\right) \end{aligned} $$ a) What is the distance of this car from the origin of the coordinate system at \(t=5.00 \mathrm{~s} ?\) b) What is the velocity vector as a function of time? c) What is the speed at \(t=5.00 \mathrm{~s} ?\)

In a projectile motion, the horizontal range and the maximum height attained by the projectile are equal. a) What is the launch angle? b) If everything else stays the same, how should the launch angle, \(\theta_{0},\) of the projectile be changed for the range to be halved?

Your friend's car is parked on a cliff overlooking the ocean on an incline that makes an angle of \(17.0^{\circ}\) below the horizontal. The brakes fail, and the car rolls from rest down the incline for a distance of \(29.0 \mathrm{~m}\) to the edge of the cliff, which is \(55.0 \mathrm{~m}\) above the ocean, and, unfortunately, continues over the edge and lands in the ocean. a) Find the car's position relative to the base of the cliff when the car lands in the ocean. b) Find the length of time the car is in the air.

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