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# 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? # 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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Other asked 2021-05-18
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?

## Answers (1) 2021-05-20
The range of a projectile that lands at the same height it waslaunched is
$$\displaystyle\triangle{x}={\frac{{{{v}_{{0}}^{{2}}}{\sin{{\left({2}\theta\right)}}}}}{{{g}}}}$$
We can solve this for $$\displaystyle\theta$$ by multiplying by g, and dividing by $$\displaystyle{{v}_{{{0}}}^{{2}}}$$:
$$\displaystyle{\sin{{\left({2}\theta\right)}}}={\frac{{{g}\triangle{x}}}{{{{v}_{{0}}^{{2}}}}}}$$
Now take the inverse sine of both sides:
$$\displaystyle{2}\theta={{\sin}^{{-{1}}}{\left({\frac{{{g}\triangle{x}}}{{{{v}_{{0}}^{{2}}}}}}\right)}}$$
Finally, divide by 2:
$$\displaystyle\theta=\frac{{{\sin}^{{-{1}}}{\left({\frac{{{g}\triangle{x}}}{{{{v}_{{0}}^{{2}}}}}}\right)}}}{{2}}$$
We are given an initial velocity of $$\displaystyle{v}_{{0}}={460}\frac{{m}}{{s}}$$, adistance of $$\displaystyle\triangle{x}={45.7}{m}$$, and the acceleration due to gravity is $$\displaystyle{g}={9.8}\frac{{m}}{{s}^{{2}}}$$, so inserting these into our expression for $$\displaystyle\theta$$, we have
θ = sin-1[(9.8 m/s)(45.7 m) /(460 m/s)] / 2 = 1.06*10^-3rad
In degrees this is
$$\displaystyle\theta={1.06}\cdot{10}^{{-{3}}}{\left(\frac{{180}}{\pi}\right)}={0.0606}^{\circ}$$

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