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

Options

A. `30^@`
B. `60^@`
C. `45^@`
D. `0^@`

Answer

B

Explanation

Maximum height `H=(u^2sin^(2) theta)/(2g)`
or `gH=(u^2sin^(2)theta)/2 " "...(i)`
Velocity at the highest point , `v_H = u cos theta`
Let `v_x, v_y` be the horizontal and vertical of projectile at hight .
`H/2 `. Then , `v_x = u cos theta` and
`v_y^2=u^2sin^(2) theta - 2g xxH/2 =u^2 sin^(2) theta- gH`
`= u^2 sin^2 theta-(u^2 sin^2 theta)/2=(u^2sin^2 theta)/2 " "` (Using (i))
`:.` Net velocity at height `H/2 = (v_x^2+v_y^2)^(1//2)`
As per question
`sqrt(2/5)(v_x^2+v_y^2)^(1//2)=v_H" or " 2/5(v_x^2+v_y^2)=v_H^2`
or `2/5[u^2cos^(2) theta +u^2/2 sin^(2) theta] =u^2 cos^(2) theta`
or `sin^(2) theta = 3 cos^(2) theta " or " sintheta = sqrt3 cos theta `
or `tan theta = sqrt3 = tan 60^(@) " or " theta = 60^@`