Question

A particle aimed at a target, projected with an angle ${15}^o$ with the horizontal is short of the target by $10m$. If projected with an angle of ${45}^o$ is away from the target by $15m$, then the traget by $15m$, then the angle of projection to hit the target is:

• A. $\frac{1}{2}si{n}^{-1}\left(\frac{1}{10}\right)$
• B. $\frac{1}{2}si{n}^{-1}\left(\frac{3}{10}\right)$
• C. $\frac{1}{2}si{n}^{-1}\left(\frac{9}{10}\right)$
• D. $\frac{1}{2}si{n}^{-1}\left(\frac{7}{10}\right)$

Answer 1

The correct option is D $\frac{1}{2}si{n}^{-1}\left(\frac{7}{10}\right)$

$R_1=R-10$

$R_2=R+15$

$\frac{R-10}{R+15}=\frac{sin2{\theta }_1}{sin2{\theta }_2}=\frac{sin{30}^o}{sin{90}^0}=\frac{1}{2}$

$2R-20=R+15$

$\Rightarrow R=35$

$R_{max}=\frac{u^2}{g}$$\Rightarrow 50=\frac{u^2}{g}$$\Rightarrow u^2=50g$

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

$35=\frac{50g\times sin2\theta }{g}$

$sin2\theta =\frac{35}{50}=\frac{7}{10}$

$2\theta =si{n}^{-1}\left[\frac{7}{10}\right]$

$\theta =\frac{1}{2}si{n}^{-1}\left[\frac{7}{10}\right]$