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Q.
The maximum range of a gun on horizontal terrain is $ 16\,km $ if $ g=10\,m/{{s}^{2}} $ . What must be the muzzle velocity of the shell?
BHUBHU 2003
Solution:
For maximum horizontal range body should be projected at an angle of $45^{\circ}$.
The horizontal range covered is given by $R=u_{x} \times T$
$=u \cos \theta \times \frac{2 u \sin \theta}{g} $
$=u^{2} \frac{(2 \sin \theta \cos \theta)}{g}$
$R=\frac{u^{2} \sin 20}{g}$
For $R_{\max } \sin 2 \theta=1$
$\therefore \theta=45^{\circ} $
$\therefore R_{\max }=\frac{u^{2}}{g} $
$\Rightarrow u=\sqrt{R_{\max } g} $
Given, $ R_{\max }=16\, km =16 \times 10^{3} \,m, g=10 \,m / s ^{2} $
$u=\sqrt{16 \times 10^{3} \times 10} $
$u=400 \,m / s$
Note: Horizontal range is same whether the body is projected at $\theta$ or $\left(90^{\circ}-\theta\right)$ i. e., for complementary angles of projection range is same.
