Q. 4.29 A bullet fired at an angle of 30^{\circ} 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)
D Devendra Khairwa

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

                                                          R\ =\ \frac{u^2\ \sin 2\Theta }{g}

or                                                       3\ =\ \frac{u^2\ \sin 60^{\circ} }{g}

or                                                        \frac{u^2}{g}\ =\ 2\sqrt{3}

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

 

                                                          R_{max}\ =\ \frac{u^2\ \sin 2(45^{\circ}) }{g}

or                                                                     =\ 3.46\ Km

Thus the bullet cannot travel up to 5 Km.

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