# 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.

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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