Question

The range \(R\) of a projectile fired with an initial velocity \(v_{0}\) at an angle \(\theta\) with the horizontal is \(R=\frac{v_{0}^{2} \sin 2 \theta}{g},\) where \(g\) is the acceleration due to gravity. Find the angle \(\theta\) such that the range is a maximum.

Short answer

The angle \(\theta\) at which the range \(R\) of the fired projectile is maximized is \(\frac{\pi}{4}\) or 45 degrees.

Step-by-step solution

Write Down the Given Equation

The range \(R\) of a projectile fired with an initial velocity \(v0\) at an angle \(\theta\) with the horizontal is represented by the following equation: \(R=\frac{{v_{0}}^{2} \sin(2\theta)}{g}\). This equation is a standard formula in physics which represents the horizontal displacement covered by the projectile.

Differentiate the Equation with Respect to Theta

To find the angle at which the range \(R\) of the projectile is maximized, let's differentiate \(R\) with respect to \(\theta\). The derivative \(\frac{dR}{d\theta}\) of \(R\) with respect to \(\theta\) becomes: \(\frac{dR}{d\theta} = \frac{{v_{0}}^{2}}{g} \cos(2\theta)\). Here, a derivative has been used because it helps find the maximum or minimum values of a function. In this case, we want to maximize \(R\), the range.

Set the Derivative to Zero

For \(R\) to be a maximum, its derivative must be equal to zero. Thus, set \(\frac{dR}{d\theta} = 0\) to find the value of \(\theta\) at which this occurs. This gives us: \(\frac{{v_{0}}^{2}}{g} \cos(2\theta) = 0\), which simplifies to \(\cos(2\theta) = 0\). The cosine function is zero at \(\frac{\pi}{2}\) plus an integral multiple of \(\pi\), but since \(\theta = \frac{1}{2} \times 2\theta\), the first value within the range 0 to \(\pi\) is \(\frac{\pi}{4}\). Therefore, the angle at which the range is maximum is \(\theta = \frac{\pi}{4}\).

Verify the Result

To confirm that this indeed gives a maximum range, it's good to show that the second derivative, \(\frac{d^{2}R}{d\theta^{2}}\), is negative at \(\theta = \frac{\pi}{4}\). The second derivative is: \(\frac{d^{2}R}{d\theta^{2}} = -\frac{{2{v_{0}}^{2}}}{g} \sin(2\theta)\). At \(\theta = \frac{\pi}{4}\), the second derivative becomes: \(-\frac{{2{v_{0}}^{2}}}{g} \sin(\frac{\pi}{2}) = -\frac{{2{v_{0}}^{2}}}{g}\), which is a negative value. Thus, \(\theta = \frac{\pi}{4}\) does indeed give a maximum range.