question_answer

A body is projected at such angle that the horizontal range is three times the greatest height. The .angle of projection is

A)  \[{{42}^{o}}8\]

B)  \[{{53}^{o}}7\]

C)  \[{{33}^{o}}7\]

D)  \[{{25}^{o}}8\]

Correct Answer: B

Solution :

Let a body be projected at a velocity u at an angle \[\theta \] with the horizontal. Then horizontal range covered is given by                 \[R=\frac{{{u}^{2}}{{\sin }^{2}}2\,\theta }{g}\] ... (i) and height H is \[H=\frac{{{u}^{2}}{{\sin }^{2}}2\,\theta }{2\,g}\] ... (ii) Given, \[R=3\,H\] \[\therefore \] \[\frac{{{u}^{2}}\sin \,2\theta }{g}=3\times \frac{{{u}^{2}}{{\sin }^{2}}\theta }{2\,g}\] Also, \[\sin \,2\theta =2\sin \theta \cos \theta \] \[\therefore \] \[\frac{{{u}^{2}}2\sin \,2\cos \theta }{g}=3\times \frac{{{u}^{2}}{{\sin }^{2}}\theta }{2\,g}\] \[\Rightarrow \] \[\tan \theta =\frac{2}{1.5}=1.33\] \[\Rightarrow \] \[\tan \theta =\frac{2}{1.5}=1.33\] \[\theta ={{53}^{o}}7\] Hence, angle of projection is \[{{53}^{o}}7\].