The velocity of a projectile when it is at the greatest height is `(sqrt (2//5))` times its velocity when it is at half of its greatest height. Determine its angle of projection.

Answer

`60^(@), tan^(-1) (sqrt(3//2))`

Explanation

For vertical component of velocity of projection
`0 = u_(y)^(2) 2gH rArr u_(y)^(2) 2gH ..(1)`
vertical component of velocity at `H//2`
`v_(y)^(2) = u_(y)^(2) - 2g xx (H)/(2) rArr v_(y)^(2) = u_(y)^(2) - gH ..(2)`
from (1) & (2)
`v_(y)^(2) = u_(y)^(2) - (u_(y)^(2))/(2) rArr v_(y)^(2) = (u_(y)^(2))/(2)`
From energy7 conservation and given condition
speed at `H = sqrt((2)/(5))` speed at `(H)/(2)`
`u_(x) = sqrt((2)/(5)) sqrt(u_(x)^(2) +v_(y)^(2)) rArr u_(x)^(2) = (2)/(5) (u_(x)^(2)+(u_(y)^(2))/(2))`
`rArr u_(x)^(2) = (1)/(5) (2u_(x)^(2) +u_(y)^(2))`
`5u_(x)^(2) = 2u_(x)^(2) +u_(y)^(2) rArr 3u_(x)^(2) = u_(y)^(2) rArr (u_(y))/(u_(x)) = sqrt(3)`
`rArr tan theta = sqrt(3) rArr theta = 60^(@)`