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.

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 3.0 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.
Answer» Solution :We are given that angle of projection with the horizontal, `theta = 30^(@)`, horizontal range R = 3km. As `R = (v_(0)^(2) sin 2 theta)/(g) , 3 = (v_(0)^(2) sin 60^(@))/(g) = (v_(0)^(2))/(g) xx (sqrt(3))/(2)` or `(v_(0)^(2))/(g) = 2sqrt(3)km` SINCE the muzzle speed `(v_(0))` is fixed, `R_("max") = (v_(0)^(2))/(g) = 2sqrt(3) = 2 xx 1.732 = 3.464 km` Obviously, it is not POSSIBLE to hit the target 5km away.