A graph between the square of the velocity of a particle and the distance(s) moved is shown in figure. The acceleration of the particle in kilometers per hour square is:-

Options

A. `(a) 2250`
B. `(b) 250`
C. `(c) -2250`
D. `(d) -250`

Answer

C

Explanation

In vertical direction `h=(u sin theta)t-1/2 g t^(2)`
`rArr t^(2)-((2u sin theta)/g)t+(2h)/g=0`
`rArr t_(1)+t_(2)=(2u sin theta)/g` …(i)
In horizontal direction `x=(u cos theta)t-1/2 at^(2)`
`rArr t^(2)-((2u cos theta)/a)t+(2x)/a=0`
`rArr t_(3)+t_(4)=(2u cos theta)/a`...(ii)
From (i) and (ii) `theta=tan^(-1)[(g(t_(1)+t_(2)))/(a(t_(3)+t_(4)))]`