A) \[R=4\frac{\sqrt{H{{H}_{1}}}}{{}}\]
B) \[R=\frac{\sqrt{H{{H}_{1}}}}{{}}\]
C) \[R=4H{{H}_{1}}\]
D) None of the above
Correct Answer: A
Solution :
Range of projectile, \[R=\frac{2{{u}^{2}}\,\,\sin \theta \,\,\cos \theta }{g}\] ?..(i) Height \[H=\frac{{{u}^{2}}\,{{\sin }^{2}}\theta }{2g}\] ?..(ii) \[{{H}_{1}}=\frac{{{u}^{2}}\,{{\sin }^{2}}\,({{90}^{o}}-\theta )}{2g}=\frac{{{u}^{2}}\,{{\cos }^{2}}\theta }{2g}\] ?..(iii) Then, \[H{{H}_{1}}=\frac{{{u}^{2}}\,{{\sin }^{2}}\theta \,{{u}^{2}}\,{{\cos }^{2}}\theta }{2g\,2g}\] ??(lv) From Eq. (i), we get \[{{R}^{2}}=\frac{4{{u}^{2}}\,si{{n}^{2}}\,\,\theta \,{{u}^{2}}\,{{\cos }^{2}}\,\theta \times 4}{2g2g}\] \[R=\sqrt{16\,\,H{{H}_{1}}}\] [from Eq. (iv)] \[=4\sqrt{\,H{{H}_{1}}}\]
You need to login to perform this action.
You will be redirected in
3 sec
