A cannon ball has a range R on a horizontal plane. If h and `h^(')` are the greatest heights in the two paths for which this is possible, then-

Options

A. `R=4sqrt(hh^('))`
B. `R=(4h)/(h^('))`
C. `R=4hh^(')`
D. `R=sqrt(hh^('))`

Answer

A

Explanation

For same range there are two possible angles of projection `thetaand 90^(@)-theta` ltBrgt Let u be velocity of projection
`thereforeR=(u^(2)sin2theta)/(g)` ltBrgt In `I^(st)` case, h = `(u^(2)sin^(2)theta)/(2g)`
In llnd case, h' = `(u^(2)sin^(2)(90-0))/(2g)=(u^(2)cos^(2)theta)/(2g)`
`therefore` hh'= `(u^(4)sin^(2)thetacos^(2)theta)/(4g^(2))`
`=(u^(4)4sinthetacos^(2)theta)/(16g^(2))`
`=((u^(2)sin2theta)/(g))^(2)=R^(2)`
`thereforeR=sqrt(16hh')=4sqrt(hh')`