Answer:
vₓ = 10 m / s, [tex]v_{y}[/tex] = 10 - 10 t
Explanation:
This is a projectile launching exercise, where the x and y axes are treated independently.
On the x axis. towards the north there is no acceleration, so the speed is constant
vₓ = v₀ₓ
On the vertical y-axis, there is the acceleration of gravity with value (-g) is directed down, the equation is
[tex]v_{y}[/tex] = [tex]v_{oy}[/tex] - gt
Let's use trigonometry to find the initial velocity component
sin 45 = [tex]v_{oy}[/tex] / v₀
cos 45 = v₀ₓ / v₀
[tex]v_{oy}[/tex] = v₀ sin45
v₀ₓ = v₀ cos45
To find the initial velocity let's use the scope equation
R = v₀² sin 2θ / g
v₀ = √ Rg / sin 2θ
v₀ = √ (20 10 / sin (2 45))
v₀ = 14.14 m / s
With this value we find the initial velocity component
[tex]v_{oy}[/tex] = 14.14 sin 45
[tex]v_{oy}[/tex] = 10 m / s
v₀ₓ = 14.14 cos 45
v₀ₓ = 10 m / s
We get the equations to graph
X axis
vₓ = 10 m / s
In this graph it is a horizontal line
Y Axis
[tex]v_{y}[/tex] = 10 - 10 t
For this graph let's build a table
t (s) [tex]v_{y}[/tex] (m / s)
0 10
0.5 5
1.0 0
1.5 -5
2 -10
see attached for both curves
