Vi = 21.1m/s @ 58.3deg.
X = hor = 21.1cos58.3 = 11.09m/s.
Y=ver = 21.1sin58.3 = 17.95m/s @ t = 0.
a. V = Vi + 0.5gt,
V = 17.95 + 0.5(-9.8)3.3,
V = 17.95 - 16.17 = 1.78m/s @ 3.3s.
An arrow is shot into the air at an angle of 58.3 above the horizontal with a speed of 21.1 m/s.
(a) What are the x and y components of the velocity of the arrow 3.3 s after it leaves the bowstring?
I found the x component to be 11.09 m/s, but I don't know how to find the y component.
1 answer
1 answer