A rifle that shoots bullets at 460 m/s is to be aimed at atarget 45.7 m away. if center of the target is level with therifle, how high above the target must the rifle barrel be pointedso that the bullet hits the dead center?

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The range of a projectile that lands at the same height it waslaunched is △x=v02sin⁡(2θ)g We can solve this for θ by multiplying by g, and dividing by v02: sin⁡(2θ)=g△xv02 Now take the inverse sine of both sides: 2θ=sin−1(g△xv02) Finally, divide by 2: θ=sin−1(g△xv02)2 We are given an initial velocity of v0=460ms, adistance of △x=45.7m, and the acceleration due to gravity is g=9.8ms2, so inserting these into our expression for θ, we have θ=sin−1[(9.8m/s)(45.7m)/(460m/s)]/2=1.06×10−3rad In degrees this is θ=1.06⋅10−3(180π)=0.0606∘

A rifle that shoots bullets at 460 m/s is to be aimed at atarget 45.7 m away. if center of the target is level with therifle, how high above the target must the rifle barrel be pointedso that the bullet hits the dead center?

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