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A bullet fired at an angle of 30^@ with ...

A bullet fired at an angle of `30^@` with the horizontal hits the ground 3.0 km away. By adjusting its angle of projection, can one hope to hit a target 5.0 km away? Assume the muzzle speed to be fixed and neglect air resistance.

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We are given that angle of projection with the horizontal, `theta=30^(@)`, horizontal range R=3 km.
`R=(u_(0)^(2)sin2 theta)/(g)` ,
`3=(u_(0)^(2) sin 2theta)/(g)=(u_(0)^(2))/(g)xx(sqrt(3))/(2)`
or `(u_(0)^(2))/(g)=2sqrt(3)` km
Since the muzzle speed `(u_(0))` is fixed
`R_("max")=(u_(0)^(2))/(g)=2sqrt(3)=2xx1.732=3.464`km
so, it is not possible to hit the target 5 km away.
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