STATEMENT -1 : For an observer looking out through the window of a fast moving train , the nearby objects appear to move in the opposite direction to the train , while the distant objects appear to be stationary .
STATEMENT - 2 : If the observer and the object are moving at velocities ` vec v_(1)` and `vec v_(2)` respecttively with refrence to a laboratory frame , the velocity of the object with respect to a laboratory frame , the velocity of the object with respect to the observer is `vecv_(2) - vecv(1) ` .
(a) Statement -1 is True, statement -2 is true , statement -2 is a correct explanation for statement -1
(b) Statement 1 is True , Statement -2 is True , statement -2 is NOT a correct explanation for statement -1
(c) Statement - 1 is True , Statement -2 is False
(d) Statement -1 is False, Statement -2 is True

Options

A. Statement I is true, Statement II is true, Statement II is the correct explanation for Statement I.
B. Statement I is true, Statement II is true, Statement II is not the correct explanation for Statement I.
C. Statement I is true, Statement II is false.
D. Statement I is false, Statement II is true.

Answer

`((a^(2))/(4b), tan^(-1) (a))`

Explanation

`((a^(2))/(4b), tan^(-1)(a))`
Given that `y=ax-bx^(2)`
Comparing it with equation of a projectile, we get
`y=tan theta -(gx^(2))/(2u^(2) cos^(2) theta) rArr tan theta=a ...(i) and (g)/(2u^(2) cos^(2) theta)=b ......(ii)`
`u^(2)=(g)/(2b cos^(2) theta)=(g(a^(2)+1))/(2b) ........(iii)`
`[therefore cos theta=(1)/sqrt(a^(2)+1)]`
Maximum height attained, `H=(u^(2) sin^(2) theta)/(2g) .....(iv)`
From Eq. (ii), `g/(2b)=u^(2) cos^(2) theta rArr (g)/(2b)=u^(2)-u^(2) sin^(2) theta rArr u^(2) sin^(2) theta=u^(2)-g/(2b)`
`rArr u^(2) sin^(2) theta=(g(a^(2)+1))/(2b)-(g)/(2b)=(ga^(2))/(2b) .....(v)`
From Eqs. (iv) and (v), `H=(ga^(2))/(2b xx 2g)=a^(2)/(4b)` and angle of projection with horizontal is `theta=tan^(-1) (a)`