Question

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 a projectile be changed for the range of the projectile to be halved?

Short answer

How should the launch angle be changed to halve the range while keeping everything else the same? Answer: The launch angle for the projectile to have an equal horizontal range and maximum height is 45°. To halve the range of the projectile while keeping everything else the same, the launch angle should be changed to 15°.

Step-by-step solution

Recall the formulas for horizontal range and maximum height

In projectile motion, the horizontal range (R) and maximum height (H_max) can be found using the following equations: \(R = \frac{v_{0}^{2}sin(2\theta)}{g}\) \(H_{max} = \frac{v_{0}^{2}sin^2(\theta)}{2g}\) where \(v_{0}\) is the initial velocity, \(\theta\) is the launch angle, and \(g\) is the acceleration due to gravity (approximately \(9.81 m/s^2\)). We are given that \(R = H_{max}\), so we can set these formulas equal to each other: \(\frac{v_{0}^{2}sin(2\theta)}{g} = \frac{v_{0}^{2}sin^2(\theta)}{2g}\)

Solve for the launch angle

To find the launch angle, \(\theta\), in part a), we will solve the equation we obtained in step 1 by canceling common terms and simplifying: \(v_{0}^{2}sin(2\theta) = 2v_{0}^{2}sin^2(\theta)\) Cancel \(v_{0}^{2}\) from both sides: \(sin(2\theta) = 2sin^2(\theta)\) Recall that \(sin(2\theta) = 2sin(\theta)cos(\theta)\). Substitute this in the equation: \(2sin(\theta)cos(\theta) = 2sin^2(\theta)\) Divide both sides by \(2sin(\theta)\): \(cos(\theta) = sin(\theta)\) Recall that \(sin(\theta) = cos(90^{\circ}-\theta)\). Therefore: \(cos(\theta) = cos(90^{\circ}-\theta)\) Thus, the angle \(\theta = 45^{\circ}\).

Answer a)

The launch angle for the projectile to have an equal horizontal range and maximum height is \(\theta = 45^{\circ}\).

Find the changed launch angle to halve the range

In part b), we need to find a new launch angle, \(\theta_{0}\), such that the horizontal range is halved while keeping everything else the same. Using the formula for the horizontal range, we have: \(\frac{1}{2}R = \frac{v_{0}^{2}sin(2\theta_{0})}{g}\) Since we know the original range \(R = \frac{v_{0}^{2}sin(2\times45^{\circ})}{g}\), substitute this into the equation: \(\frac{1}{2} \cdot \frac{v_{0}^{2}sin(2\times45^{\circ})}{g} = \frac{v_{0}^{2}sin(2\theta_{0})}{g}\) Simplify and cancel the common terms: \(sin(2\times45^{\circ}) = 2sin(2\theta_{0})\) Now \(sin(2\times45^{\circ}) = sin(90^{\circ}) = 1\). So we have: \(1 = 2sin(2\theta_{0})\) Divide both sides by 2: \(sin(2\theta_{0}) = \frac{1}{2}\) Find the angle: \(2\theta_{0} = 30^{\circ}\) Finally, divide both sides by 2: \(\theta_{0} = 15^{\circ}\)

Answer b)

To halve the range of the projectile while keeping everything else the same, the launch angle should be changed to \(\theta_{0} = 15^{\circ}\).

Key concepts

Launch Angle

The launch angle of a projectile is the angle at which it is launched above the horizontal axis. It's one of the most critical factors affecting the projectile's trajectory, determining the distance (range) and the highest point (maximum height) the projectile will achieve during flight.

In the context of our exercise, a 45-degree launch angle is optimal when we aim to maximize both range and height equally. This is due to the symmetrical nature of the projectile's path when projected at this angle under ideal conditions—without air resistance. A fascinating aspect of a 45-degree launch angle is that it results in equal horizontal range and maximum height, provided the initial velocity is kept constant.

Horizontal Range

The horizontal range of a projectile refers to the total horizontal distance traveled from the launch point to the landing point. This aspect is inherently linked to the launch angle; a higher angle tends to yield a higher maximum height but can decrease the horizontal range.

In a scenario where a projectile's horizontal range and maximum height are equal, the formula \( R = \frac{v_{0}^{2}\sin(2\theta)}{g} \) simplifies under the condition that the launch angle is at 45 degrees. This specific launch angle balances the vertical and horizontal components of motion to achieve the equal values for range and height.

Maximum Height

Maximum height in projectile motion is the highest vertical point the projectile reaches in its trajectory. The formula \( H_{\text{max}} = \frac{v_{0}^{2}\sin^2(\theta)}{2g} \) showcases how the vertical component of the initial velocity and the launch angle influence the peak of the projectile's flight.

The example given in the exercise demonstrates the direct relationship between launch angle and maximum height. If the projectile's horizontal range is equal to its maximum height, the calculations suggest the ideal launch angle is 45 degrees. However, this scenario assumes a vacuum where air resistance is absent, which would otherwise impact the maximum height reached.

Initial Velocity

Initial velocity is the speed at which a projectile is launched. It's a vector quantity, meaning it has both magnitude and direction. The vector's horizontal and vertical components contribute respectively to the distance traveled and the maximum height achieved.

From our exercise, we observe that initial velocity serves as a common factor in both equations for range and maximum height. For a given mass and launch angle, it is the initial velocity that singularly determines how far and how high the projectile will go. Altering any one aspect of a projectile's launch—be it the angle, the velocity, or both—can have significant effects on its trajectory and final landing position.