Question

The equation of motion of a projectile is $ y = ax - b{x^2} $ where $ a $ and $ b $ are constants of motion. Match the quantities in column-I with the relations in column-II.

Column I	Column II
(A) The initial velocity of projection.	(p) $ a/b $
(B) The horizontal range of projectile (q) $ a\sqrt {\dfrac{2}{{bg}}} $ and	(q) $ a\sqrt
(C) The maximum vertical height attained by projectile	(r) $ {a^2}/4b $
(D) Time of flight of projectile (s) $ \sqrt {\dfrac{{g(1 + {a^2})}}{{2b}}} $	(s) $ \sqrt

(A) A-p, B-q, C-r, D-s
(B) A-s, B-p, C-q, D-r
(C) A-s, B-p, C-r, D-q
(D) A-p, B-s, C-r, D-q

Answer 1

Hint: The question provides us with an equation for the motion of a projectile, which is also the equation of a parabola. By comparing this to the equation of a projectile, we can relate the velocity and the angle of projection to the given constants. Which in turn gives the derived formulas in column II.

Formula used
 $ y = x\tan \theta - \dfrac{{g{x^2}}}{{2{u^2}{{\cos }^2}\theta }} $

Complete Step by step solution
We know that the equation of the projectile is,
 $ y = x\tan \theta - \dfrac{{g{x^2}}}{{2{u^2}{{\cos }^2}\theta }} $
where $ x $ is the coordinate of the X-axis,
 $ y $ is the coordinate of the Y-axis,
 $ g $ is the acceleration due to gravity,
 $ u $ is the projection velocity of the object
and $ \theta $ is the angle of projection of the object.
In the question, the equation of the projectile is,
 $ y = ax - b{x^2} $
Comparing both of these equations, we find that the coefficients,
 $ a = \tan \theta $
and $ b = \dfrac{g}{{2{u^2}{{\cos }^2}\theta }} $
Now deriving the equations given in column I,
(A) Initial velocity of projection,
  For a projectile, the initial velocity is denoted by $ u $ .
You might notice that $ b $ contains the term $ u $ , so we can change $ b $ such that the other variable $ \theta $ also gets replaced with a constant.
Writing the value of $ b $ ,
 $ b = \dfrac{g}{{2{u^2}{{\cos }^2}\theta }} $
Writing $ {\cos ^2}x $ as $ \dfrac{1}{{{{\sec }^2}x}} $ ,
 $ \Rightarrow b = \dfrac{{g{{\sec }^2}\theta }}{{2{u^2}}} $
Using Identity, $ {\sec ^2}x = 1 + {\tan ^2}x $ ,
 $ \Rightarrow b = \dfrac{{g(1 + {{\tan }^2}\theta )}}{{2{u^2}}} $
Substituting $ a = \tan \theta $ ,
 $ \Rightarrow b = \dfrac{{g(1 + {a^2})}}{{2{u^2}}} $
Rearranging the equation,
 $ \Rightarrow u = \sqrt {\dfrac{{g(1 + {a^2})}}{{2b}}} $
Therefore, (A) from column I matches with (s) from column II.
(B) The horizontal range of projectile
The range $ R $ of a projectile is given by,
 $ R = \dfrac{{{u^2}\sin 2\theta }}{g} $
Using the identity, $ \sin 2\theta = 2\sin \theta \cos \theta $ ,
 $ \Rightarrow R = \dfrac{{{u^2}2\sin \theta \cos \theta }}{g} $
Multiplying numerator and denominator with $ \cos \theta $ ,
 $ \Rightarrow R = \dfrac{{{u^2}2\sin \theta \cos \theta }}{g} \times \dfrac{{\cos \theta }}{{\cos \theta }} $
 $ \Rightarrow R = \dfrac{{{u^2}2}}{g} \times \dfrac{{\sin \theta }}{{\cos \theta }} \times {\cos ^2}\theta $
Converting the trigonometric identities,
 $ \Rightarrow R = \dfrac{{2{u^2}}}{g} \times \tan \theta \times \dfrac{1}{{{{\sec }^2}\theta }} $
 $ \Rightarrow R = \dfrac{{2{u^2}}}{g} \times \tan \theta \times \dfrac{1}{{(1 + {{\tan }^2}\theta )}} $
Replacing,
 $ a = \tan \theta $
and the value of $ u $ in terms of $ a $ and $ b $ ,
 $ \Rightarrow R = \dfrac{2}{g} \times a \times \dfrac{1}{{(1 + {a^2})}} \times \dfrac{{g(1 + {a^2})}}{{2b}} $
Simplifying,
 $ \Rightarrow R = \dfrac{a}{b} $
Therefore, (B) from column I matches with (p) from column II.
(C) The maximum vertical height attained by projectile
The maximum vertical height attained by the projectile is given by,
 $ H = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}} $
Multiplying numerator and denominator with $ {\cos ^2}\theta $ ,
 $ H = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}} \times \dfrac{{{{\cos }^2}\theta }}{{{{\cos }^2}\theta }} $
Using trigonometric identities,
 $ H = \dfrac{{{u^2}{{\tan }^2}\theta }}{{2g}} \times \dfrac{1}{{{{\sec }^2}\theta }} $
 $ \Rightarrow H = \dfrac{{{u^2}{{\tan }^2}\theta }}{{2g}} \times \dfrac{1}{{(1 + {{\tan }^2}\theta )}} $
Replacing,
 $ a = \tan \theta $
And the value of $ u $ in terms of $ a $ and $ b $ ,
 $ \Rightarrow H = \dfrac{{{u^2}{a^2}}}{{2g}} \times \dfrac{1}{{(1 + {a^2})}} $
 $ \Rightarrow H = \dfrac{{{a^2}}}{{2g}} \times \dfrac{1}{{(1 + {a^2})}} \times \dfrac{{g(1 + {a^2})}}{{2b}} $
Simplifying,
 $ \Rightarrow H = \dfrac{{{a^2}}}{{4b}} $
Therefore, (C) from column I matches with (r) from column II.
(D) Time of flight of projectile
The time of flight of a projectile is given by,
 $ T = \dfrac{{2u\sin \theta }}{g} $
Multiplying and dividing this equation with $ \cos \theta $ .
 $ T = \dfrac{{2u\sin \theta }}{g} \times \dfrac{{\cos \theta }}{{\cos \theta }} $
Solving the trigonometric identities,
 $ T = \dfrac{{2u\cos \theta }}{g} \times \tan \theta $
We know that,
 $ b = \dfrac{g}{{2{u^2}{{\cos }^2}\theta }} $
Rearranging this value as
 $ {u^2}{\cos ^2}\theta = \dfrac{g}{{2b}} $
 $ \Rightarrow u\cos \theta = \sqrt {\dfrac{g}{{2b}}} $
Using this value in the formula for $ T $ ,
 $ T = \dfrac{2}{g} \times \tan \theta \times \sqrt {\dfrac{g}{{2b}}} $
Replacing $ \tan \theta = a $
 $ T = \dfrac{2}{g} \times a \times \sqrt {\dfrac{g}{{2b}}} $
Simplifying,
 $ T = a\sqrt {\dfrac{2}{{gb}}} $
Therefore, (D) from column I matches with (q) from column II.
The correct combination for the match is, A-s, B-p, C-r, D-q, which makes option (3) the correct answer.

Note
The two variables of the projectile motion, that are the velocity of the body and the angle of projection, must be replaced with the given constants, $ a $ and $ b $ . Only these values must be removed, other quantities which are constant like numerical coefficients or $ g $ can be left in the equation.