Let's break down the problem step by step: --- ## **Problem Restatement** A projectile is fired vertically from point **A** with an initial speed of \( v_ = 220 \) ft/sec. An observer is at **B**, 485 ft horizontally from **A**. At what two times \( t_1 \) and \( t_2 \) will the line of sight from **B** to the projectile make an angle of \( 22^\circ \) with the horizontal? Also, find the speed of the projectile at these times, accounting for **drag**: - Drag deceleration: \( k v^2 \), \( k = 2. \times 10^{-3} \) ft\(^{-1}\) - Gravity: \( g = 32.2 \) ft/s² --- ## **Step 1: Find the Height Where Angle is \( 22^\circ \) from B** Let \( y \) be the height above point A at time \( t \). From the geometry: \[ \tan(22^\circ) = \frac{y}{485} \implies y = 485 \times \tan(22^\circ) \] Calculate: \[ \tan(22^\circ) \approx .404 \] \[ y = 485 \times .404 \approx 195.94\ \text{ft} \] --- ## **Step 2: Equation of Motion with Drag** The vertical motion equation (upward is positive): \[ m \frac{dv}{dt} = -mg - k v^2 \] Divide both sides by \( m \): \[ \frac{dv}{dt} = -g - k v^2 \] ### **(a) Upward:** \[ \frac{dv}{dt} = -g - k v^2 \] This is a separable ODE. ### **(b) Downward:** Direction reverses, but drag always opposes motion. (Same equation, but sign of \( v \) reverses.) --- ## **Step 3: Upward Motion -- Solve for \( v(t) \) and \( y(t) \)** This is a classic drag + gravity problem. ### **A. Terminal velocity:** Set \( \frac{dv}{dt} = \): \[ = -g - k v_{term}^2 \implies v_{term}^2 = -\frac{g}{k} \] But this is negative (since up is positive, terminal velocity is negative for falling). For upward motion, use direct integration. ### **B. Solve for \( v(t) \):** \[ \frac{dv}{dt} = -g - k v^2 \implies \frac{dv}{g + k v^2} = -dt \] Let’s solve: \[ \frac{1}{g + k v^2} = \frac{1}{g} \cdot \frac{1}{1 + \frac{k}{g} v^2} \] Let \( \alpha^2 = \frac{k}{g} \implies \alpha = \sqrt{\frac{k}{g}} \): \[ \frac{dv}{g(1 + \alpha^2 v^2)} = -dt \] \[ \int \frac{dv}{1 + \alpha^2 v^2} = -g \int dt \] \[ \frac{1}{\alpha} \tan^{-1}(\alpha v) = -g t + C \] Solve for \( v \): \[ \tan^{-1}(\alpha v) = -g \alpha t + C_1 \] \[ \alpha v = \tan(-g \alpha t + C_1) \] \[ v(t) = \frac{1}{\alpha} \tan(-g \alpha t + C_1) \] Initial condition: at \( t = , v = v_ \): \[ v_ = \frac{1}{\alpha} \tan(C_1) \implies C_1 = \tan^{-1}(\alpha v_) \] So, \[ v(t) = \frac{1}{\alpha} \tan\left[\tan^{-1}(\alpha v_) - g \alpha t\right] \] --- ### **Step 4: Find Time(s) When Height \( y = 195.94 \) ft** The height \( y \) as a function of time: \[ y(t) = \int_^t v(t) dt \] Let’s write the integral using the above \( v(t) \): \[ y(t) = \int_^t \frac{1}{\alpha} \tan\left[\tan^{-1}(\alpha v_) - g\alpha t'\right] dt' \] Let \( \theta_ = \tan^{-1}(\alpha v_) \). Let’s use substitution: \( u = \theta_ - g\alpha t' \implies du = -g\alpha dt' \implies dt' = -\frac{du}{g\alpha} \) Change limits: when \( t' = , u = \theta_ \); when \( t' = t, u = \theta_ - g\alpha t \) So: \[ y(t) = \int_{t'=}^{t'=t} \frac{1}{\alpha} \tan\left[\theta_ - g\alpha t'\right] dt' = \frac{1}{\alpha} \int_{u=\theta_}^{u=\theta_ - g\alpha t} \tan(u) \cdot \left(-\frac{du}{g\alpha}\right) \] \[ = -\frac{1}{g\alpha^2} \int_{\theta_}^{\theta_ - g\alpha t} \tan(u) du \] \[ = -\frac{1}{g\alpha^2} \left[ -\ln|\cos(u)| \right]_{\theta_}^{\theta_ - g\alpha t} \] \[ = \frac{1}{g\alpha^2} \left( \ln|\cos(\theta_)| - \ln|\cos(\theta_ - g\alpha t)| \right) \] \[ = \frac{1}{g\alpha^2} \ln \left| \frac{\cos(\theta_)}{\cos(\theta_ - g\alpha t)} \right| \] Set \( y(t) = 195.94 \) ft: \[ 195.94 = \frac{1}{g\alpha^2} \ln \left| \frac{\cos(\theta_)}{\cos(\theta_ - g\alpha t)} \right| \] --- ## **Step 5: Plug in the Numbers** First, compute \( \alpha \): \[ \alpha = \sqrt{\frac{k}{g}} = \sqrt{\frac{2. \times 10^{-3}}{32.2}} \] \[ = \sqrt{6.211 \times 10^{-5}} \] \[ = 7.885 \times 10^{-3}\ \text{(rounded)} \] Now, \( v_ = 220 \) ft/s. \[ \alpha v_ = 7.885 \times 10^{-3} \times 220 \approx 1.735 \] \[ \theta_ = \tan^{-1}(1.735) \approx 60.1^\circ = 1.049 \text{ radians} \] \( g = 32.2 \) ft/s². Now, \[ 1/(g\alpha^2) = 1 / (32.2 \times (7.885 \times 10^{-3})^2) = 1 / (32.2 \times 6.22 \times 10^{-5}) = 1 / (.002004) \approx 499 \] So: \[ 195.94 = 499 \cdot \ln\left|\frac{\cos(1.049)}{\cos(1.049 - 32.2 \times 7.885 \times 10^{-3} \times t)}\right| \] \[ \frac{195.94}{499} = \ln\left|\frac{\cos(1.049)}{\cos(1.049 - .254 t)}\right| \] \[ .3926 = \ln\left|\frac{\cos(1.049)}{\cos(1.049 - .254 t)}\right| \] \[ e^{.3926} = \frac{\cos(1.049)}{\cos(1.049 - .254 t)} \] \[ 1.481 = \frac{\cos(1.049)}{\cos(1.049 - .254 t)} \] \[ \cos(1.049 - .254 t) = \frac{\cos(1.049)}{1.481} \] \[ \cos(1.049) \approx .497 \] \[ \frac{.497}{1.481} \approx .336 \] So, \[ \cos(1.049 - .254 t) = .336 \] \[ 1.049 - .254 t = \arccos(.336) \] \[ \arccos(.336) \approx 1.228 \text{ radians} \] So, \[ 1.049 - .254 t_1 = 1.228 \implies -.254 t_1 = 1.228 - 1.049 = .179 \implies t_1 = -.179 / .254 = -.705 \ \text{(not physical, ignore)} \] But cosine is positive in the first and fourth quadrants, so solution also at \[ 1.049 - .254 t_2 = 2\pi - 1.228 = 5.055 \implies -.254 t_2 = 5.055 - 1.049 = 4.006 \implies t_2 = -4.006 / .254 = -15.77 \ \text{(not physical)} \] But this suggests a calculation issue. Let's check: Actually, for cosine, the equation: \[ 1.049 - .254 t = \pm 1.228 + 2\pi n \] So, \[ 1.049 - .254 t = 1.228 \implies t_1 = (1.049 - 1.228) / -.254 = .179 / .254 = .705 \ \text{seconds} \] Similarly, \[ 1.049 - .254 t = -1.228 \implies t_2 = (1.049 + 1.228) / .254 = 2.277 / .254 = 8.97 \ \text{seconds} \] --- ## **Step 6: Compute Speeds at Those Times** Recall (from earlier): \[ v(t) = \frac{1}{\alpha} \tan[\tan^{-1}(\alpha v_) - g \alpha t] \] Where \( \alpha = .007885 \), \( \tan^{-1}(\alpha v_) = 1.049 \), \( g\alpha = .254 \). ### **For \( t_1 = .705 \) s:** \[ v_1 = \frac{1}{.007885} \tan(1.049 - .254 \times .705) = 126.8 \tan(1.049 - .179) = 126.8 \tan(.87) \] \[ \tan(.87) \approx 1.19 \] \[ v_1 = 126.8 \times 1.19 = 150. \text{ ft/s} \] ### **For \( t_2 = 8.97 \) s:** \[ v_2 = \frac{1}{.007885} \tan(1.049 - .254 \times 8.97) = 126.8 \tan(1.049 - 2.277) = 126.8 \tan(-1.228) \] \[ \tan(-1.228) \approx -2.27 \] \[ v_2 = 126.8 \times (-2.27) = -288 \text{ ft/s} \] --- ## **Step 7: Final Answers** ### **A. Times:** \[ \boxed{t_1 = .71\ \text{s} \quad \text{(on the way up)}} \] \[ \boxed{t_2 = 9.\ \text{s} \quad \text{(on the way down)}} \] ### **B. Speeds:** \[ \boxed{v_1 = 150\ \text{ft/s}\ \text{(upward)}} \] \[ \boxed{v_2 = 288\ \text{ft/s}\ \text{(downward)}} \] --- ## **Summary Table** | Time (s) | Speed (ft/s) | Direction | |----------|--------------|-------------| |