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 the projectile be changed for the range to be halved?

Short answer

Based on the given information where the horizontal range and maximum height are equal, we found that: a) The initial launch angle of the projectile is \(45^\circ\). b) To halve the horizontal range while keeping everything else the same, the launch angle should be changed to \(15^\circ\).

Step-by-step solution

Define the equations for projectile motion

In order to solve this problem, we need to first recall the equations for projectile motion: For horizontal range \(R\) and initial velocity \(v_0\), we have \(R=\frac{v_0^2\sin(2\theta_0)}{g}\), where \(g\) is the acceleration due to gravity, and \(\theta_0\) is the launch angle. For maximum height \(H\), \(H=\frac{v_0^2\sin^2(\theta_0)}{2g}\). Now, we are given that the horizontal range and maximum height are equal, ie, \(R=H\).

Determine the launch angle

Let's equate the expressions for \(R\) and \(H\) and solve for \(\theta_0\): \(\frac{v_0^2\sin(2\theta_0)}{g} = \frac{v_0^2\sin^2(\theta_0)}{2g}\). Simplify the equation by cancelling out \(v_0^2\) and \(g\): \(\sin(2\theta_0) = 2\sin^2(\theta_0)\). Recall the double-angle identity for sine: \(\sin(2\theta_0) = 2\sin(\theta_0)\cos(\theta_0)\). Substituting into the equation and cancelling out the \(\sin(\theta_0)\) term, we are left with: \(\cos(\theta_0) = \sin(\theta_0)\). Divide both sides by \(\cos(\theta_0)\): \(\tan(\theta_0) = 1\). So, the launch angle is: \(\theta_0 = \arctan(1) = 45^\circ\).

Determine the new launch angle to halve the range

Now, we need to find the new launch angle, say \(\phi\), such that the new range, say \(R'\) is half of the original range: \(R' = \frac{1}{2}R\). Substitute the expressions for \(R\) and \(R'\) in terms of the initial velocity and launch angles, and divide both sides by the initial \(R\) to get: \(\frac{\sin(2\phi)}{\sin(2\theta_0)} = \frac{1}{2}\). We know that \(2\theta_0 = 45^\circ \times 2 = 90^\circ\), so \(\frac{\sin(2\phi)}{\sin(90^\circ)} = \frac{1}{2}\). Now, we can solve for \(2\phi\): \(\sin(2\phi) = \frac{1}{2}\), \(2\phi = \arcsin(\frac{1}{2})\), \(2\phi = 30^\circ\). Finally, the new launch angle is: \(\phi = \frac{2\phi}{2} = 15^\circ\).

Answer

a) The launch angle is \(45^\circ\). b) The launch angle should be changed to \(15^\circ\) for the range to be halved.

Key concepts

Launch Angle

Understanding the launch angle is critical in predicting the trajectory of a projectile. The launch angle is the angle at which an object is projected into the air relative to the horizontal ground. It is denoted by \(\theta_{0}\) and plays a significant role in determining the path and distance a projectile will travel.

In our exercise, we found that when the horizontal range and maximum height are equal, the launch angle is \(45^\circ\). This specific angle is significant—it's where the range is maximized for a given initial velocity in the absence of air resistance. Furthermore, any adjustments to the launch angle can greatly alter the projectile's trajectory. For example, halving the range required the launch angle to be changed to \(15^\circ\).

This manipulation of the launch angle is important in many real-world applications, from sports to ballistic trajectories. By mastering how the launch angle affects motion, one gains better control over predicting and controlling the path of the projectile.

Maximum Height

The maximum height a projectile reaches is another crucial aspect of its trajectory. It refers to the highest vertical point the projectile achieves in its flight before gravity pulls it back down. This concept in projectile motion is influenced by the initial velocity and the launch angle.

In the given problem, the formula \(H=\frac{v_0^2\sin^2(\theta_0)}{2g}\) expresses the relationship between these factors. When dealing with maximum height, only the vertical component of the initial velocity—given by \(v_0\sin(\theta_0)\)—and the acceleration due to gravity (\(g\)) are considered.

Understanding this relationship can help predict how different angles and velocities influence the height a projectile reaches. It's essential not just for homework problems, but for real-world planning, like determining how high a fountain should spray water to achieve a desired aesthetic appearance.

Horizontal Range

Finally, the horizontal range of a projectile is the distance it travels along the horizontal axis from the point of launch to where it first contacts the ground. The horizontal range is dependent on both the launch angle and the initial velocity, as captured by the formula \(R=\frac{v_0^2\sin(2\theta_0)}{g}\).

It is crucial to realize that at the optimal angle of \(45^\circ\), the range achieves its maximum value because \(\sin(90^\circ)=1\), which is the maximum value for the sine function. The exercise demonstrates that altering the launch angle from \(45^\circ\) to \(15^\circ\) effectively halves the range. This understanding is fundamental in numerous activities such as golf, where knowing how to alter your swing to change the range of the ball can be a game-changer.

In any field that requires an object to be sent flying—the greater the comprehension of how the horizontal range works, the better the application and execution of skills that rely on it.