Question

HELP!!!

An arrow is shot into the air at an angle of 30.0 above the horizontal with a speed of 20.0 m/s. What are the x and y components of the
velocity of the arrow 1.0 s after it leaves the bowstring?

Answer 1

Y(1s) =
`10√(3)` - 10.1

X(1s) = 10m/s

Step-by-step 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:

`Yo =(√(3) )/(2) . Vr\\Yo = (√(3) )/(2) . 20\\Yo = 10√(3) m/s \\Xo = (1)/(2) Vr\\Xo = 10m/s`

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 =
`10√(3)` - 10.1

If you prefere, can be: Yf = 10. (
`√(3) - 1`)