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.
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.
To find the y-component of the velocity, you can use the formula:
V_y = V_initial * sin(theta)
Where:
V_y is the y-component of the velocity
V_initial is the initial velocity of the arrow (21.1 m/s in this case)
theta is the angle of the arrow above the horizontal (58.3 degrees in this case)
Plug in the values into the formula:
V_y = 21.1 m/s * sin(58.3 degrees)
Using a calculator:
V_y ≈ 18.04 m/s
Therefore, the y-component of the velocity of the arrow 3.3 seconds after it leaves the bowstring is approximately 18.04 m/s.