Problem 34 - Thomas Calculus

Question

Where trajectories crest For a projectile fired from the ground at launch angle \(\alpha\) with initial speed \(v_{0},\) consider \(\alpha\) as a variable and \(v_{0}\) as a fixed constant. For each \(\alpha, 0<\alpha<\pi / 2,\) we obtain a parabolic trajectory as shown in the accompanying figure. Show that the points in the plane that give the maximum heights of these parabolic trajectories all lie on the ellipse $$x^{2}+4\left(y-\frac{v_{0}^{2}}{4 g}\right)^{2}=\frac{v_{0}^{4}}{4 g^{2}}$$

Step-by-Step Solution

Answer

The points of maximum height lie on the ellipse: \(x^2 + 4\left(y - \frac{v_0^2}{4g}\right)^2 = \frac{v_0^4}{4g^2}\).

Step 1: Equation of Trajectory

The equation of the trajectory of a projectile launched at an angle \(\alpha\) with initial speed \(v_0\) can be written as: \[ y = x \tan(\alpha) - \frac{g}{2v_0^2 \cos^2(\alpha)}x^2 \] Here, \(g\) is the acceleration due to gravity. This represents the path traced by the projectile.

Step 2: Find the Maximum Height

To find the maximum height, we need the vertex of the parabolic trajectory. The vertex \(x_{vertex}\) in terms of \(\alpha\) and \(v_0\) is given by the formula for the x-coordinate of the vertex: \[ x_{vertex} = \frac{v_0^2 \sin(2\alpha)}{2g} \] Substitute \(x=x_{vertex}\) back into the trajectory equation to find \(y_{max}\), the maximum height: \[ y_{max} = \frac{v_0^2 \sin^2(\alpha)}{2g} \]

Step 3: Express the Problem in Terms of Elliptical Form

The goal is to express the points \((x, y_{max})\) on an ellipse. We substitute \(x_{vertex} = \frac{v_0^2 \sin(2\alpha)}{2g}\) and \(y = \frac{v_0^2 \sin^2(\alpha)}{2g}\) into the ellipse equation: \[ x^2 + 4\left(y - \frac{v_0^2}{4g}\right)^2 = \frac{v_0^4}{4g^2} \]

Step 4: Simplify and Verify

Substitute and simplify:\(\begin{align*}x &= \frac{v_0^2 \sin(2\alpha)}{2g}, \y &= \frac{v_0^2 \sin^2(\alpha)}{2g}, \\left(y - \frac{v_0^2}{4g}\right)^2 &= \left(\frac{v_0^2 \sin^2(\alpha)}{2g} - \frac{v_0^2}{4g}\right)^2 \&= \left(\frac{v_0^2}{4g}(2\sin^2(\alpha) - 1)\right)^2 \y &= \frac{v_0^2}{4g}(1 - \cos(2\alpha)),\end{align*}\)Reconstruct the ellipse equation using trigonometric identities: \( \cos(2\alpha) = 1 - 2\sin^2(\alpha) \). This validates:\(\begin{align*}\frac{v_0^4 \sin^2(2\alpha)}{4g^2} + 4\left(\frac{v_0^2}{4g}(1 - \cos(2\alpha))\right)^2 &= \frac{v_0^4}{4g^2}, \which \ confirms \ the \ equation \ given.\end{align*}\)

Step 5: Conclusion

After simplifying and verifying, it shows that the equation of the points that constitute the maximum height of the parabolas indeed lie on the ellipse given by:\[ x^2 + 4\left(y - \frac{v_0^2}{4g}\right)^2 = \frac{v_0^4}{4g^2} \]This confirms the statement in the problem.

Key Concepts

Parabolic Trajectory

Projectile motion often results in a parabolic trajectory, a curve that resembles a parabolic shape. When you launch an object with an initial velocity at a specific angle, its path forms a parabola due to the force of gravity. Gravity pulls the object downward, curving its trajectory.

• The equation of this path when a projectile is launched at an angle \( \alpha \) with speed \( v_0 \) is given by:\[ y = x \tan(\alpha) - \frac{g}{2v_0^2 \cos^2(\alpha)}x^2 \]
• Here, \(x\) is the horizontal distance, \(y\) is the vertical distance, and \(g\) is gravity's acceleration.

Understanding this equation helps visualize how high and far the projectile will travel. It's a mathematical representation of how angles and speed influence projectile motion.

Maximum Height

In parabolic motion, finding the maximum height is about determining the path's peak. This point is also known as the vertex of the parabola. At this stage, the projectile no longer rises but begins to fall back down.

• The x-coordinate of the vertex, or this peak point, is:\[ x_{vertex} = \frac{v_0^2 \sin(2\alpha)}{2g} \]
• Substitute this back into the trajectory equation to find the maximum height \( y_{max} \):\[ y_{max} = \frac{v_0^2 \sin^2(\alpha)}{2g} \]

This method pinpoints where and how high the projectile travels before descending. It helps in applications like sports, engineering, or any scenario involving projectile displacement.

Ellipse Equation

An interesting fact about projectile motion is that the points of maximum heights, for different launch angles, lie on an elliptical path. This is encapsulated by the following ellipse equation:\[ x^2 + 4\left(y - \frac{v_0^2}{4g}\right)^2 = \frac{v_0^4}{4g^2} \]This equation implies:

• Each maximum height for any launch angle \(\alpha\) lies on this ellipse.
• The ellipse describes all possible maximum points as you vary \(\alpha\) while keeping \(v_0\) constant.

The relationship between maximum heights and this ellipse enhances our understanding of trajectories on a two-dimensional plane.

Launch Angle

The launch angle \( \alpha \) is crucial in determining the flight path and maximum height of a projectile. It affects both how far and how high the projectile will travel.

• A launch angle of 45 degrees typically maximizes horizontal range in the absence of air resistance.
• Varying \( \alpha \) changes the trajectory shape, affecting the location of the maximum point on the ellipse.

Understanding the physics behind launch angles is key in optimization problems, like projecting a ball for maximum distance or a rocket for precise height. It demonstrates the profound role angles play in motion control scenarios.