Projectiles why the person multiplied the column vector OB by h?

[phospho]

Question 7c on attached,

in the solutions could someone explain why the person multiplied the column vector OB by h? I don't get it

 

[Delphi51]

I'm glad you said it was an "h"; looked like a "u" to me which would really be mysterious.
He is just saying the velocity vector must be a constant times the displacement vector OB. Same as saying the ratio of the vertical to the horizontal must be the same for the displacement and the velocity in order for them to have the same slope and be parallel. The h just takes care of the units so h*(50 meters) = 10 m/s
Funny, the h doesn't seem to be used in the following steps. I would have followed up the
velocity vector = [Vx,Vy] = h[50, -52.5]
with [10, 14 - gt] = [50h, -52.5*50h]
and then 10 = 50h so h = 0.2 and 14 - gt = -52.5*50h
which solves to t = 2.5.

Alternatively, slope of OB = = slope of trajectory at time t
Dy/Dx = Vy/Vx
-52.5/50= (14- gt)/10
-525/50 = 14 - gt
gt = 14 +525/50 = 24.5
t = 24.5/g = 2.50.

 

[voko]

The velocity vector must be parallel to OB.

 

[phospho]

I'm glad you said it was an "h"; looked like a "u" to me which would really be mysterious.
He is just saying the velocity vector must be a constant times the displacement vector OB. Same as saying the ratio of the vertical to the horizontal must be the same for the displacement and the velocity in order for them to have the same slope and be parallel. The h just takes care of the units so h*(50 meters) = 10 m/s
Funny, the h doesn't seem to be used in the following steps. I would have followed up the
velocity vector = [Vx,Vy] = h[50, -52.5]
with [10, 14 - gt] = [50h, -52.5*50h]
and then 10 = 50h so h = 0.2 and 14 - gt = -52.5*50h
which solves to t = 2.5.

Alternatively, slope of OB = = slope of trajectory at time t
Dy/Dx = Vy/Vx
-52.5/50= (14- gt)/10
-525/50 = 14 - gt
gt = 14 +525/50 = 24.5
t = 24.5/g = 2.50.

cheers