1. The problem statement, all variables and given/known data A low-flying helicopter flying a constant 215 km/h horizontally wants to drop secret documents to his contact's open car which is traveling 155 km/h on a level highway 78.0 m below. At what angle (to the horizontal) should the car be in his sights when the packet is released? 2. Relevant equations Kinematic Equations for Constant Acceleration in Two Dimensions X-component (horizontal): vx = vx0 + axt x = x0 + vx0t + .5axt2 vx2 = vx02 + 2ax(x-x0) Y-Component (vertical): vy = vy0 + ayt y = y0 + vy0t + .5ayt2 vy2 = vy02 + 2ay(y-y0) 3. The attempt at a solution Knowns: Vhelicopter = 59.7 m/s Vcar = 43.1 m/s y = -78 m y0 = 0 x0 = 0 g = -9.8 m/s2 vy2 = ? vy2 = vy02 + 2ay(y-y0) vy2 = 0 + 2 (-9.8 m/s2)(-78 m) vy = 39.1 m/s vx = Vhelicopter + Vcar = 59.7 m/s + 43.1 m/s = 102.8 m/s ? <-- I'm not sure I did this right. Probably didn't. tan Θ = vy/vx = 39.1/102.8 Θ = 20 degrees But the back of the book says that the answer is 49.6 degrees. I'm probably wrong and somewhere in my work I screwed up somewhere. My calculator is in degree mode.