Dear student 
Taking positive x axis parallel to OB and y axis perpendicular to it (downward as positive)
 
Initial condition
under initial condition
ux = ucos 45
uy = usin 45o 
ax = -gsin45
ay = gcos45

to determine t
using equation of motion along x axis
vx = ux +axt
as vx = 0 
so t = -(ux/ax)
or, t =  ucos 45/gsin45​
t = u/g
this is the time of flight
now using it
we drop a perpendicular from projection point on OB
​the length of the projection is l = 10 sin45​
 
using equation of motion along y axis
y = uy + ayt^2/2
10sin 45 = usin45 + 12gcos 45(u/g)2
we get 10 = u + u^2/2g
or u^2 + 2ug = 20g
or, u^2 +20u - 200 = 0
solving this quadratic equation we get,
u = 7.32 m/s 
​
 
at point B ,  it has only velocity in y direction so its velocity in x direction has become zero
hence using the equation of motion along y axis
vx^2 = ux^2 + 2ax

as vx = 0

so x = -(u cos45)2 / 2(-gsin 45)
x = u^2 /2sqrt(2)
x = 1.89 m 
so using equation of motion along y axis
vy = uy +ayt
vy = usin 45 + u(gcos45)/g
vy = usin45 + ucos45
thus total velocity at B is vy as vx = 0
v = 7.32 * sqrt(2)  = 10.35 m /s