Delphi51 said: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.
The person likely multiplied the column vector OB by h to calculate the vertical displacement of the projectile. By multiplying the magnitude of OB by h, they can determine how far the projectile has traveled vertically.
The column vector OB represents the initial velocity of the projectile. It contains both the magnitude and direction of the initial velocity, which is essential in determining the trajectory of the projectile.
Multiplying the column vector OB by h allows for the calculation of the vertical displacement of the projectile. This information, along with the initial velocity, can help determine the height, range, and flight time of the projectile.
Yes, it is essential to consider the vertical displacement in projectile motion. The vertical displacement affects the trajectory and flight time of the projectile and is necessary to accurately calculate the overall motion of the projectile.
Yes, the column vector OB can be multiplied by any value, as long as it is used consistently throughout the calculation. For example, it could be multiplied by time to determine the horizontal displacement of the projectile.