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Answer to Question #10465 in Mechanics | Relativity 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.

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Comments

Assignment Expert
30.11.16, 18:53

Dear Fred, please use panel for submitting new questions.

Fred
25.11.16, 19:47

I do not understand the answer.

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