Question

The trajectory of a projectile in a vertical plane isy = ax – bx²,where a and b are constants and x and y are respectively horizontal and vertical distances of the projectile from the point of projection. The maximum height attained by the particle and the angle of projection from the horizontal are:

• A. $\frac{b^{2}}{2a},\tan^{- 1}(b)$
• B. $\frac{a^{2}}{b},\tan^{- 1}(2a)$
• C. $\frac{a^{2}}{4b},\tan^{- 1}(a)$
• D. $\frac{2a^{2}}{b},\tan^{- 1}(a)$

Answer 1

Answer: (C)

$\frac{a^{2}}{4b},\tan^{- 1}(a)$

Answer 2

Q y = ax – bx²

a = tan q and $\frac { g } { 2 u ^ { 2 } \cos ^ { 2 } \theta } = b$

and $\frac { u ^ { 2 } \sin ^ { 2 } \theta } { 2 g }$ = H

or $\frac { \mathrm { g } } { 2 \mathrm { u } ^ { 2 } \cos ^ { 2 } \theta } \times \frac { \mathrm { u } ^ { 2 } \sin ^ { 2 } \theta } { 2 \mathrm {~g} } = \mathrm { Hb }$

or $\frac { \tan ^ { 2 } \theta } { 4 } = \mathrm { Hb }$ or H = $\frac { \mathrm { a } ^ { 2 } } { 4 \mathrm {~b} }$ and q = tan^–1(1)