# Answer on Mechanics | Relativity Question for kiva lamoure

Question #10465

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?

Expert's answer

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.

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.

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## Comments

Assignment Expert30.11.2016 11:53Dear Fred, please use panel for submitting new questions.

Fred25.11.2016 12:47I do not understand the answer.

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