A body is thrown at an angle `theta_0` with the horizontal such that it attains a speed equal to `sqrt((2)/(3))` times the speed of projection when the body is at half of its maximum height. Find the angle `theta_0`.

Options

A. `15^@`
B. `30^@`
C. `45^@`
D. `60^@`

Answer

B

Explanation

(b) `v_y^2 - v^2 sin^2 = -2g [(1)/(2) (v^2 sin^2 theta)/(2g)]`
or `v_y^2 = v^2 sin^2 theta - (v^2 sin^2 theta)/(2) or v_y = (v sin theta)/(sqrt(2))`
`v'^2 = (v cos theta)^2 + ((v sin theta)/(sqrt(2)))^2`
=`v^2 cos^2 theta + (v^2 sin^2 theta)/(2) = v^2 (cos^2 theta+ (sin^2 theta)/(2))`
`v'= v sqrt(cos^2 theta + (sin^2 theta)/(2))`
`v^2 cos^2 theta = (2)/(5) v^2[cos^2 theta + (sin^2 theta)/(2)]`
or `5 cos^2 theta = 2 cos^2 theta + sin^2 theta or 3 cos^2 theta = sin^2 theta`
or `tan^2 theta = 3 or tan theta = sqrt(3) or theta = 60^@`
Angle of projection with vertical is `90^@ - 60^@ = 30^@`.