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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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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