Answer to Question #10465 in Mechanics | Relativity for kiva lamoure

Question

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?

Answer 1

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.