An airplane flies at a constant height h and constant speed j. A gun is at the origin (0,0). A shell is fired from the gun with an initial muzzle velocity v at an angle theta with the horizontal. What

You're on a roll — The correct option is C. v = (j^2 + 2gh)^1/2. Step 1: Define the motion of the airplane and the shell. Let the gun be at the origin (0,0). The airplane flies at a constant height h and constant speed j. At time t, its position is (x_A, y_A) = (jt, h). The shell is fired with an initial muzzle velocity v at an angle with the horizontal. At time t, its position is (x_S, y_S). The horizontal component of the shell's initial velocity is v_x = v . The vertical component of the shell's initial velocity is v_y = v . The equations of motion for the shell are: x_S(t) = (v ) t y_S(t) = (v ) t - (1)/(2) g t^2 Step 2: Set the conditions for the shell to hit the airplane. For the shell to hit the airplane at time t, their positions must be equal: x_S(t) = x_A(t) (v ) t = j t Since t ≠ 0 (the shell takes time to reach the airplane), we can divide by t: v = j (*) y_S(t) = y_A(t) (v ) t - (1)/(2) g t^2 = h (**) Step 3: Express in terms of v and j. From (*), = (j)/(v). Using the identity ^2 + ^2 = 1: = sqrt(1 - ^2) = sqrt(1 - ((j)/(v))^2) = sqrt((v^2 - j^2)/(v^2)) = sqrt(v^2 - j^2)v Step 4: Substitute into equation (**) and solve for t. Substitute into (**): v (v