## College Physics (4th Edition)

(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 $7.86~m$
(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 $7.86~m$