A) 4 hh
B) 8 hh
C) 2 hh
D) 16 hh
E) 24 hh
Correct Answer: D
Solution :
The range is given by \[R=\frac{{{u}^{2}}\sin 2\theta }{g}\] ?(i) Also we know that for pair of angles \[\theta \]\[90-\theta ,\] and the range will remain same. Now, heights attained by projectiles \[h=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{g}\] and \[h=\frac{{{u}^{2}}\sin \,(90-\theta )}{2g}\] \[\Rightarrow \] \[hh=\frac{{{u}^{4}}{{\sin }^{2}}\theta \,{{\cos }^{2}}\theta }{4{{g}^{2}}}\] \[\Rightarrow \] \[hh=\frac{{{u}^{2}}2\sin \theta \cos \theta }{16{{g}^{2}}}\] \[\Rightarrow \] \[hh={{\left( \frac{4\sin 2\theta }{g} \right)}^{2}}\times \frac{1}{16}\] ?(ii) From Eqs. (i) and (ii) we get \[16\,hh={{R}^{2}}\]
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