Question

A projectile is launched twice from a height \(y_{0}=0\) at a given launch speed, \(v_{0} .\) The first launch angle is \(30.0^{\circ} ;\) the second angle is \(60.0^{\circ} .\) What can you say about the range \(R\) of the projectile in these two cases? a) \(R\) is the same for both cases. b) \(R\) is larger for a launch angle of \(30.0^{\circ}\). c) \(R\) is larger for a launch angle of \(60.0^{\circ}\). d) None of the preceding statements is true.

Short answer

Answer: a) The range is the same for both cases.

Step-by-step solution

Calculate the Range for the First Launch Angle of \(30.0^{\circ}\)

First, we will calculate the range for the first launch angle of \(30.0^{\circ}\). By plugging the given angle and velocity into the range formula, we have: \(R_1 = \frac{v_{0}^2\sin 2(30.0^{\circ})}{g}\). We don't need to compute the range's exact value since we're only comparing to the other launch angle. Keep the formula for the range for later reference.

Calculate the Range for the Second Launch Angle of \(60.0^{\circ}\)

Next, we will calculate the range for the second launch angle of \(60.0^{\circ}\). Plug the given angle and velocity into the range formula: \(R_2 = \frac{v_{0}^2\sin 2(60.0^{\circ})}{g}\). Once again, we don't need to compute the range's exact value, as we are just comparing with the other launch angle. Keep the formula for the range.

Compare the Ranges

Now we can compare the two expressions: \(R_1 = \frac{v_{0}^2\sin 2(30.0^{\circ})}{g} = \frac{v_{0}^2\sin (60.0^{\circ})}{g}\) and \(R_2 = \frac{v_{0}^2\sin 2(60.0^{\circ})}{g} = \frac{v_{0}^2\sin (120.0^{\circ})}{g}\). Notice that \(\sin (120.0^{\circ})=\sin (180.0^{\circ}-60.0^{\circ})=\sin(60.0^{\circ})\). Since \(\sin60^{\circ} = \sin120^{\circ}\), then \(R_1 = R_2\). From the comparison, we can conclude: a) \(R\) is the same for both cases. So, the correct answer is option a.

Key concepts

Launch Angle Physics

Understanding the physics behind the launch angle is crucial when examining the motion of a projectile. The launch angle is the angle between the initial velocity vector and the horizontal plane. This angle greatly affects the trajectory and the range of the projectile. Kinematic equations dictate that the maximum range of a projectile is achieved at a 45-degree launch angle under ideal conditions.

However, the range is the same for launch angles that add up to 90 degrees, a concept known as the complementary angle theorem. In our exercise, the launch angles of 30 degrees and 60 degrees are complementary angles (they add up to 90 degrees), leading to the same range for the projectile assuming that the launch speed is constant and air resistance is negligible. This phenomenon can be explained by the symmetry in the sine function used to calculate the range where \(\sin(\theta) = \sin(90^\circ - \theta)\), hence \(\sin(2 \cdot 30^\circ) = \sin(2 \cdot 60^\circ)\), when the angle is measured in degrees.

Trajectory Comparison

When comparing trajectories, it's vital to consider both the launch angle and the initial velocity. The same initial speed with different launch angles can result in differing trajectory shapes and thereby affect the landing point or range of the projectile. Trajectories will be parabolic in nature, but the height and distance traveled will vary.

For any given speed, a launch angle of 45 degrees offers the ideal compromise between vertical and horizontal distances, resulting in the farthest range. The angles less than 45 degrees will give a flatter trajectory and are typically used when the aim is to hit targets closer to the ground level. Conversely, angles greater than 45 degrees will result in a steeper climb and fall, useful when trying to launch over obstacles. Despite these differences, complementary angles (those that add to 90 degrees) will produce the same range, facilitating a versatile approach to achieving the same distance.

Kinematics Equations

Projectile motion is governed by kinematics equations, which reveal how objects move under the influence of gravity assuming a near-earth surface and ignoring air resistance. The fundamental equations relate variables such as initial velocity, time, displacement, acceleration due to gravity (\(g\)), and the angle of projection.

The range of a projectile can be found using the equation \( R = \frac{v_{0}^2\sin(2\theta)}{g} \) where \( v_{0} \) is the launch speed, \( \theta \) is the launch angle, and \( g \) is the acceleration due to gravity. This equation derives from the horizontal and vertical components of the projectile's motion being independently analyzed and then combined to give the total distance along the horizontal axis when the object lands back at its original height. For our textbook problem, we applied this equation to show that the ranges for launch angles of 30 and 60 degrees are identical when initial speeds are the same.