Asked by Unknown3
The range of a target is found to be 20km.A shell leaves again with a velocity of 500ms-1.what must be the angle of elevation of the gun.if the ground is leveled.
Answers
Answered by
bobpursley
20km=timeinair*500*cosAngle
solve for time in air.
in the vertical
hf=hi+vi*timeinair-1/2 *(9.8)timeinair^2
or time in air=500*sinAngle/4.9
solve now for the angle.
solve for time in air.
in the vertical
hf=hi+vi*timeinair-1/2 *(9.8)timeinair^2
or time in air=500*sinAngle/4.9
solve now for the angle.
Answered by
henry2,
Range = Vo^2*sin(2A)/g = 20,000 m.
500^2*sin(2A)/9.8 = 20.000,
2A =
A =
500^2*sin(2A)/9.8 = 20.000,
2A =
A =
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