1.

A particle is projeted with a velocity v, so that its range on a horizontal plane is twice the greatest height attained. If g is acceleration due to gravity, then its range is :A. `(4v^(2))/(5g)`B. `(4g)/(5v^(2))`C. `(4v^(3))/(5g^(2))`D. `(4v)/(5g^(2))`

Answer» Correct Answer - a
`H=(v^(2) sin^(2) theta)/(2g) and R=(v^(2) sin 2 theta)/g`
Since `R=2H, (v^(2) sin 2 theta)/g=2xx(v^(2) sin^(2) theta)/(2g)`
or `2 sin theta cos theta= sin^(2) theta or tan theta=2`
`R=v^(2)xx2/gxxsin theta cos theta`
`=(2v^(2))/g =2/(sqrtg)xx1/(sqrt5)=(4v^(2))/(5g)`


Discussion

No Comment Found

Related InterviewSolutions