A bullet fired at an angle of 30^(@) with the horizontal hits the ground 3.0km 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.

A bullet fired at an angle of 30^(@) with the horizontal hits the grou

1.	A bullet fired at an angle of 30^(@) with the horizontal hits the ground 3.0km 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.
Answer» <html><body><p></p>Solution :We are <a href="https://interviewquestions.tuteehub.com/tag/given-473447" style="font-weight:bold;" target="_blank" title="Click to know more about GIVEN">GIVEN</a> that angle of projection with the horizontal, `theta=30^(@)`, horizontal range `R = 3km`. <br/> As `R=(v_(0)^(2)<a href="https://interviewquestions.tuteehub.com/tag/sin2theta-3032597" style="font-weight:bold;" target="_blank" title="Click to know more about SIN2THETA">SIN2THETA</a>)/(g),<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>=(v_(0)^(2)sin60^(0))/(g)=(v_(0)^(2))/(g)xx(sqrt(3))/(2)` <br/> or `(v_(0)^(2))/(g)=2sqrt(3)km` <br/> <a href="https://interviewquestions.tuteehub.com/tag/since-644476" style="font-weight:bold;" target="_blank" title="Click to know more about SINCE">SINCE</a> the muzzle speed `(v_(0))` is fixed, <br/>`R_(max)=(v_(0)^(2))/(g)=2sqrt(3)=2xx(1.732=3.464km` <br/> Obviously, it is not possible to hit the target 5km away.</body></html>