* Show that in the limit of strong damping (large \(\gamma\) ) the time of flight of a projectile (on level ground) is approximately \(t \approx(w / g+1 / \gamma)(1-\) \(\left.\mathrm{e}^{-1-\gamma w / g}\right)\). Show that to the same order of accuracy the range is \(x \approx\) \((u / \gamma)\left(1-e^{-1-\gamma w / g}\right) .\) For a projectile launched at \(800 \mathrm{~ms}^{-1}\) with \(\gamma=\) \(0.1 \mathrm{~s}^{-1}\), estimate the range for launch angles of \(30^{\circ}, 20^{\circ}\) and \(10^{\circ}\).

We are given the formula to calculate the time of flight (t) under strong damping: \[t \approx \frac{w}{g} + \frac{1}{\gamma}(1-e^{-(1 + \frac{\gamma w}{g})})\]

We are given the formula to calculate the range (x) under strong damping: \[x \approx \frac{u}{\gamma}\left(1-e^{-(1+\frac{\gamma w}{g})}\right)\]

For each launch angle \(\theta\), we will calculate the corresponding angular velocity (\(w\)) and their horizontal (\(u_x\)) and vertical (\(u_y\)) velocity components as follows: \[w = u\sin\theta\] \[u_x = u\cos\theta\] \[u_y = u\sin\theta\]

For each launch angle, we will calculate the time of flight (t) and the range (x) using the formulas from Steps 1 and 2, with the given values: \(u = 800\) m/s and \(\gamma = 0.1\) s\(^{-1}\). - For the case of 30 degrees (\(\theta = \frac{\pi}{6}\)): \[w = (800\,\text{m/s})\sin\frac{\pi}{6}\] \[u_x = (800\,\text{m/s})\cos\frac{\pi}{6}\] \[u_y = (800\,\text{m/s})\sin\frac{\pi}{6}\] - For the case of 20 degrees (\(\theta = \frac{\pi}{9}\)): \[w = (800\,\text{m/s})\sin\frac{\pi}{9}\] \[u_x = (800\,\text{m/s})\cos\frac{\pi}{9}\] \[u_y = (800\,\text{m/s})\sin\frac{\pi}{9}\] - For the case of 10 degrees (\(\theta = \frac{\pi}{18}\)): \[w = (800\,\text{m/s})\sin\frac{\pi}{18}\] \[u_x = (800\,\text{m/s})\cos\frac{\pi}{18}\] \[u_y = (800\,\text{m/s})\sin\frac{\pi}{18}\] Now plug all the values in the time of flight and range formulas, and calculate the results for each launch angle. You will get the ranges for each angle.