range of a target is found to be 20km. A shell leaves a gun with a velocity of 500m/s. What must be the angle of elevation of the gun if the ground is level?
The horizontal component of the shell should be: Vhor = 500 * sinA. The vertical one: Vver = 500 * cosA. So, the time to get the highest point: 500 * cosA = g * t, so t = 500 * cos A / g. The range passed: S = Vhor * 2 * t = 500 * sinA * 2 * 500 * cos A / g = 20000 sinA * cos A = 20000 * 9.81 / (500 * 2 * 500) = 0.3924 sinA * cos A = 0.5 * sin2A = 0.3924 sin2A = 0.7848 2A = 51.7022 A = 25.8511 degrees.