[Applied maths] projectile motion

(a)

consider direction perpendicular to OA,
velocity at A = -v sin(theta) (no displacement, same but opposite to velocity at O )
Then, by equation "v=u+at", we have
-v sin(theta) = v sin(theta) - g cos(alpha) t
Thus we get the traveling time t = 2v sin(theta) / g cos(alpha)

we have constant horizontal speed = v cos(theta + alpha)
horizontal distance traveled = 2a cos(alpha)

With constant speed x time = distance traveled
part (a) can be shown

(b)
from (a), the traveling time t = 2v sin(theta) / g cos(alpha)
Consider the direction along OA,
velocity at A = 0
Then, also by equation "v=u+at", we have
0 = v cos(theta) - g sin(alpha) t
Thus we get the traveling time t = v cos(theta) / g sin(alpha)
the two time obtained are the same,
we have 2 sin(theta) / cos(alpha) = cos(theta) / sin(alpha)
So, we get 2 tan(theta) tan(alpha) = 1

Consider the direction along OA again,
By equation "v^2 = u^2 + 2as", we have
0 = v^2 [cos(theta)]^2 - 4ga sin(alpha)
Thus, v^2 = 4ga sin(alpha) / [cos(theta)]^2

now consider the direction perpendicular to OA,
when the velocity parallel to OA, velocity in this direction = 0,
By equation "v^2 = u^2 + 2as", we have
0 = v^2 [sin(theta)]^2 - 2ga cos(alpha)
Sub v^2 = 4ga sin(alpha) / [cos(theta)]^2 in this equation,
we have 2[tan(theta)]^2 tan(alpha) = 1
together with 2 tan(theta) tan(alpha) = 1 obtained,
tan(theta) = 1
so theta = pi/4