1. The muzzle speed for a Lee-Enﬁeld riﬂe is 630 meters per second. Suppose you ﬁre this riﬂe at a target 700 meters away and at the same level as the riﬂe. a - In order to hit the target, you must aim the barrel at a point above the target. How many meters above the target must you aim? Pretend there is no air resistance. $$v_0 = 630$$ $$delta\x = 700$$ $$ g = 9.81$$ 2. $$v_x = v_0 cos\theta$$ $$v_y = v_0 sin\theta - gt$$ $$y = v_0y t - 1/2gt^2$$ 3. The point at which the bullet reaches its maximum height, ##t_(stop)##, has a vertical velocity component of 0. As such, ##v_y - gt = 0##. Solving for ##t_(stop)## yields ##t_(stop) = v_y / g##. Since at its maximum height the bullet has finished 1/2 of its flight, we may assume ##t_(flight) = 2(t_(stop)) = 2v_y / g##. Keeping this equation in mind, we then try to find the angle theta for the bullet. $$x_(max) = (2v_0^2(sin\theta)(cos\theta))/g$$ $$700 = ((630^2)/(g)) (sin2\theta) [Sin\theta (cos\theta) = sin2\theta]$$ $$\theta = (arcsin(gd/v_o^2))/2$$ $$\theta = 0.49568$$ Knowing theta, we can calculate the flight time. $$t_flight = 2v_y / g = 2(630sin(.49568)/9.81 = 1.111$$ To check my work, I plugged the relevant information back into the equation to find the y value. Since y = 0 when it hits the target, then I should receive a value of 0 with the relevant information. $$y = v_0y (t) - 1/2gt^2$$ $$y = (630sin(.49568))(1.111) - 1/2(9.81)(1.111^2)$$ $$y = -3.983 x 10^-5$$ But it doesn't come out to zero. Here I used the rounded calculations, but while doing the actual work, I stored the information (using my TI-83) and used the exact values. It's really close to 0, but according to this, it would still miss the target at the time given. I kept stopping originally because the theta value seems WAY too small - you're practically aiming straight forward! I'd love some help, since I've been working on this problem for what feels like hours. Please walk through it in as much detail as possible, because it feels like I've approached the problem every which way!