question_answer

An object is thrown along a direction making an angle \[{{45}^{o}}\] with the horizontal direction. The horizontal range of the object is equal to:

A)  twice the vertical height

B)  vertical height

C)  four times the vertical height

D)  three times the vertical height

Correct Answer: C

Solution :

 Let \[u\]be the velocity of projection and 6 is angle of projection, then \[R=\frac{{{u}^{2}}\sin 2\theta }{g}\] and \[H=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{g}\] Given \[\theta ={{45}^{o}}\] \[\therefore \] \[R=\frac{{{u}^{2}}\sin {{90}^{o}}}{g}=\frac{{{u}^{2}}}{g}\] ?(i) \[H=\frac{{{u}^{2}}{{\sin }^{2}}{{45}^{o}}}{g}=\frac{{{u}^{2}}}{4g}\] From Eqs. (i) and (ii), we get