A projectile thrown with velocity v making angle `theta` with vertical gains maximum height H in the time for which the projectile remains in air, the time period is

Options

A. `sqrt(H cos theta//g)`
B. `sqrt(2H cos theta//g)`
C. `sqrt(4H//g)`
D. `sqrt(8H//g)`

Answer

D

Explanation

Max. height =`H=(v^2 sin^2 (90-theta))/(2g)` …(i)
Time of flight , `T=(2v sin (90-theta))/g`…(ii)
From (i), `(v cos theta)/g=sqrt((2H)/g)`, From (ii),
`T=2sqrt((2H)/g) =sqrt((8H)/g)`