Answer
(a) $3.0$ seconds after the arrow leaves the bowstring, the x-component of the velocity is $10.0~m/s$ and the y-component of the velocity is $-12.1~m/s$
(b) During the $3.0~s$ interval after the arrow leaves the bowstring, the x-component of the displacement is $30.0~m$ and the y-component of the displacement is $7.86~m$
Work Step by Step
(a) We can find the x-component of the velocity:
$v_x = v_0~cos~\theta = (20.0~m/s)~cos~60.0^{\circ} = 10.0~m/s$
We can find the y-component of the velocity after 3.0 seconds:
$v_{fy} = v_{0y}+a_y~t$
$v_{fy} = (20.0~m/s)~sin~60.0^{\circ}+(-9.80~m/s^2)(3.0~s)$
$v_{fy} = -12.1~m/s$
$3.0$ seconds after the arrow leaves the bowstring, the x-component of the velocity is $10.0~m/s$ and the y-component of the velocity is $-12.1~m/s$
(b) We can find the x-component of the displacement:
$x = v_x~t = (10.0~m/s)(3.0~s) = 30.0~m$
We can find the y-component of the displacement:
$y = v_{0y}~t+\frac{1}{2}a_y~t^2$
$y = (20.0~m/s)~sin~60.0^{\circ}~(3.0~s)+\frac{1}{2}(-9.80~m/s^2)(3.0~s)^2$
$y = 7.86~m$
During the $3.0~s$ interval after the arrow leaves the bowstring, the x-component of the displacement is $30.0~m$ and the y-component of the displacement is $7.86~m$
