13 R is the horizontal range of a p...
Question

# 13 R is the horizontal range of a projectile, fired with a certain speed at a certain angle with the horizontal, on a horizontal plane and h is the maximum height attained by it. Then the maximum horizontal range that can be attained with the same speed of projection as before is - (A) 2h (B) (C)2R+ 8h R- 32 (D) 2h+ R 8h 8R

JEE/Engineering Exams
Physics
Solution
238
4.0 (1 ratings)
( begin{aligned} R &=frac{2 u^{2} sin theta cos theta}{g}=frac{u^{2} sin 2 theta}{g} frac{d R}{d theta} &=frac{d u^{2}}{g} frac{d(sin 2 theta)}{d theta}=frac{2 u^{2} cos 2 theta}{g}=0 Rightarrow & 2 theta=frac{pi}{2} Rightarrow mid theta=frac{pi}{4} end{aligned} ) R will be maximum when ( theta=x / 4 ) ( frac{R}{H}=frac{4}{tan theta} ) ( Rightarrow R=frac{4 H}{tan (pi / 4)}=4 H ) ( R=frac{u^{2}}{g} ; quad H=frac{u^{2} sin ^{2}(pi / 4)}{2 g}=frac{u^{2}}{4 g} ) ( frac{R^{2}}{8 h}=frac{(4 n)^{2}}{84}=frac{16 H^{2}}{84}=24 eq R ) ( 2 R+frac{h^{2}}{8 R}=2 R+frac{H^{2}}{32 H}+pi ) ( 2 H+R^{2}=2 H+frac{(44)^{2}}{84}=2 H+frac{16 H^{2}}{84}=4 H=R ) ( Rightarrow quad 0 ) prion D) ( 2 h+frac{R^{2}}{8 h} ) is correct