Answer:
Y(1s) = [tex]10\sqrt{3}[/tex] - 10.1
X(1s) = 10m/s
Explanation:
In annex I've done the explanation for the equations that I will just present here.
Assuming that the arrow stars from the position (0 ; 0) in the Cartesian Graphic, and with Xo and Yo the initial speeds:
[tex]Yo =\frac{\sqrt{3} }{2} . Vr\\Yo = \frac{\sqrt{3} }{2} . 20\\Yo = 10\sqrt{3} m/s \\Xo = \frac{1}{2} Vr\\Xo = 10m/s[/tex]
Ignoring friction with air, Xo = Xf
So, Xo is the same during all the movement.
X(1s) = 10m/s
For Yo is different. That component is suferring reductions from gravity.
We can find Yo(1s) with one the basic functions of cinematics:
Vf = Vo + at
Vf = Final Velocity
Vo = Start Velocity
a = aceleration - gravity (g) is negative here
t = time
Yf = Yo + gt
Yf = [tex]10\sqrt{3}[/tex] - 10.1
If you prefere, can be: Yf = 10. ([tex]\sqrt{3} - 1[/tex])
