Question

The range \(R\) of a projectile is defined as the magnitude of the horizontal displacement of the projectile when it returns to its original altitude. (In other words, the range is the distance between the launch point and the impact point on flat ground.) A projectile is launched at \(t=0\) with initial speed \(v_{i}\) at an angle \(\theta\) above the horizontal. (a) Find the time \(t\) at which the projectile returns to its original altitude. (b) Show that the range is \(R=\frac{v_{\mathrm{i}}^{2} \sin 2 \theta}{g}\) [Hint: Use the trigonometric identity $\sin 2 \theta=2 \sin \theta \cos \theta .]$ (c) What value of \(\theta\) gives the maximum range? What is this maximum range?

Short answer

Answer: The time t at which a projectile returns to its original altitude is given by \(t = \frac{2 v_{i} \sin \theta}{g}\), where \(v_i\) is the initial velocity, \(\theta\) is the launch angle, and \(g\) is the acceleration due to gravity. The maximum range of the projectile is achieved when the launch angle is \(45^{\circ}\), and the maximum range is given by \(\frac{v_{i}^2}{g}\).

Step-by-step solution

Find horizontal and vertical position equations

To find the time when the projectile returns to its original altitude, we first need the equations for horizontal and vertical positions as functions of time. The equations for the horizontal and vertical positions of a projectile are given by: \[x(t) = v_{i} \cos \theta \cdot t\] \[y(t) = v_{i} \sin \theta \cdot t - \frac{1}{2} g t^{2}\] Here, \(x(t)\) and \(y(t)\) are the horizontal and vertical positions of the projectile at time \(t\), respectively, \(v_i\) is the initial velocity, \(\theta\) is the launch angle, and \(g\) is the acceleration due to gravity.

Find time to return to the initial altitude

To find the time when the projectile returns to its original altitude, we need to consider when the vertical position \(y(t)\) is equal to 0. This is because the projectile is at the same altitude as the launch point when \(y(t) = 0\). So, we should solve the equation for \(t\) when \(y(t) = 0\): \[0 = v_{i} \sin \theta \cdot t - \frac{1}{2} g t^{2}\] Rearranging the equation, we obtain: \[t = \frac{2 v_{i} \sin \theta}{g}\]

Find the range

Now that we have the time when the projectile returns to its original altitude, we can find the range of the projectile by substituting this time in the horizontal position equation: \[R = x(t) = v_{i} \cos \theta \cdot \frac{2 v_{i} \sin \theta}{g}\] Using the trigonometric identity \(\sin 2 \theta = 2\sin \theta \cos \theta\), we can simplify the range equation to: \[R = \frac{v_{i}^2\sin 2\theta}{g}\]

Determine the maximum range

To maximize the range, we need to find the angle \(\theta\) that maximizes the equation for the range. The range is maximized when the \(\sin 2\theta\) term within the equation is maximized. Since the maximum value of the sine function is 1, the maximum range occurs when: \[\sin 2\theta = 1\] Therefore, the maximum range occurs when \(\theta = \frac{\pi}{4}\) or \(\theta = 45^{\circ}\).

Find the maximum range

Finally, we can find the maximum range by substituting the optimal angle \(\theta = \frac{\pi}{4}\) into the range equation: \[R_\mathrm{max} = \frac{v_{i}^2 \sin 2\left(\frac{\pi}{4}\right)}{g} = \frac{v_{i}^2}{g}\] So, the maximum range is achieved when the launch angle is \(45^{\circ}\), and the maximum range is given by \(\frac{v_{i}^2}{g}\).