A bullet is fired from the bottom of the inclined plane at angle `theta = 37^@` with the inclined plane. The angle of incline is `30^@` with the horizontal. Find the
(a) position of the maximum height of the bullet from the inclined plane ,
(b) time of light ,
( c) range along the incline ,
(d) the value of `theta` at which the range will be maximum ,
( e) maximum range.

Explanation

(a) Taking axis system as shown in (Fig. S5.11)
At highest point, `v_y = 0`. Therefore,
`v_y^2 = u_y^2 + 2 a_y y`
`rArr 0 = (30)^2 - 2 g cos 30^@ y`
`y = 30 sqrt(3) m` (maximum height)
(b) Again foe x coordiante,
`v_y = u_y + a_y t`
`0 = 30 - g cos 30^@ xx t rArr t = 2 sqrt(3) s`
Time of flight `T = 2 xx 2 sqrt(3) s`
( c) Range, `x = u_x t + (1)/(2) a_x t^2`
=`40 xx 4 sqrt(3) - (1)/(2) g sin 30^@ xx (sqrt(3)^2)`
=`40 (4 sqrt(3) - 3) m`
(d) `theta = (pi)/(4) - (30^@)/(2) = 45^@ - 15^@ = 30^@`
( e) `(u^2)/(g(1 + sin beta)) = (50 xx 50)/(10(1 + (1)/(2))) = (2500)/(15) = (500)/(3) m`.
.