An arrow is shot into the air at an angle of 59.4° above the horizontal with a speed of 20.0 m/s. What is the x-component of the velocity of the arrow 3.30 s after it leaves the bowstring?

Question

What is the y-component of the velocity of the arrow 3.30 s after it leaves the bowstring?
What is the x-component of the displacement of the arrow during the 3.30-s interval?
What is the y-component of the displacement of the arrow during the 3.30-s interval?

Answer 1

Vo = 20m/s[59.4o]
Xo = 20*cos59.4 = 10.18 m/s.
Yo = 20*sin59.4 = 17.21 m/s.

X = Xo = 10.18 m/s. Does not change.

Y = Yo+g*t = 17.21 - 9.8*3.30=-15.13 m/s
Downward.

Answer 2

Dx = Xo*T = 10.18 * 3.30 = 33.6 m.

h = (Y^2-Yo^2)/2g
h = (-15.13)^2-(17.21^2)/-19.6 =
( 228.92-296.18)/-19.6 = 3.43 m. = Y.