Equation of a projectile is given by `y = Px - Qx^(2)` , where P and Q are constants. The ratio of maximum height to the range of the projectile is

Options

A. `(Q^(2))/(2P)`
B. `(P^(2))/(Q)`
C. 4P
D. `(P)/(4)`

Answer

D

Explanation

Given
`y = Px - Ox^(2) " ". . . (i)`
Wen know that,
`y = x tan theta - (1)/(2) g (x^(2))/(u^(2)cos^(2)theta) " "(ii) `
After comparing Eqs. (i) and (ii) ,we get
`P = tan theta " ". . . (iii)`
`Q = (g)/(2u^(2) cos^(2) theta)" ". . . (iv) `
We know that
`H = (u^(2) sin^(2) theta)/(2g) " ". . . (v)`
`R = (u^(2) sin 2 theta)/(g) " ". . . (vi)`
Dividing Eq. (v) by Eq. (vi) , we get
`(H)/(R) = (u^(2) sin^(2) theta)/(2g) xx (g)/(u^(2) sin 2 theta)`
`(H)/(R) = (sin^(2) theta)/(2 xx2 sin theta cos theta)`
`(H)/(R) = (1)/(4) xx tan theta`
from Eq. (iii) , we have
`(H)/(R) = (P)/(4)`