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.

Answers (1)

The range of bullet is given to be:-       R = 3 Km.

                                                          R = fracu^2 sin 2Theta g

or                                                       3 = fracu^2 sin 60^circ g

or                                                        fracu^2g = 2sqrt3

Now, we will find the maximum range (maximum range occurs when the angle of projection is 450).

 

                                                          R_max = fracu^2 sin 2(45^circ) g

or                                                                     = 3.46 Km

Thus the bullet cannot travel up to 5 Km.

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