Question

A projectile is shot from a gun at an angle of ${60}^o$ to the horizontal has a range of $200m$.By what angle should the gun be raised towards the vertical so that the range is $120m$?

• A. $\frac{1}{2}{\sin }^{-1}\frac{3\sqrt{3}}{10}-{60}^o$ towards the vertical
• B. $\frac{1}{2}{\sin }^{-1}3\sqrt{\frac{3}{2}}-{60}^o$ towards the vertical
• C. ${60}^o-\frac{1}{2}{\sin }^{-1}\frac{3\sqrt{3}}{10}$ away from the vertical
• D. ${30}^o$ towards the vertical

Answer 1

Answer: (C)

${60}^o-\frac{1}{2}{\sin }^{-1}\frac{3\sqrt{3}}{10}$ away from the vertical

$R=\frac{u^{2}sin2\theta }{g}$

$\frac{R_1}{R_2}=\frac{\sin 2{\theta }_1}{\sin 2{\theta }_2}$

$\frac{200}{120}=\frac{\sin 120}{\sin 2{\theta }_2}$

$\sin 2{\theta }_2=0.6\sin {120}^{\circ }$

${\theta }_2=\frac{1}{2}{\sin }^{-1}\left(0.6\sin 120\right)$
${\theta }_2=\frac{1}{2}{\sin }^{-1}\left(\frac{3\sqrt{3}}{10}\right)$