Question

required information
an arrow is shot into the air at an angle of 60.0° above the horizontal with a speed of 20.7 m/s.
what is the y - component of the velocity of the arrow 3.0 s after it leaves the bowstring?
v_y = \boxed{} m/s

Explanation:

Step1: Find initial vertical velocity

$v_{0y} = v_0 \sin\theta = 20.7 \times \sin(60.0^\circ)$
$v_{0y} = 20.7 \times \frac{\sqrt{3}}{2} \approx 17.94 \, \text{m/s}$

Step2: Apply vertical velocity kinematic equation

$v_y = v_{0y} - gt$
Use $g = 9.81 \, \text{m/s}^2$, $t=3.0 \, \text{s}$
$v_y = 17.94 - (9.81 \times 3.0)$

Step3: Calculate final vertical velocity

$v_y = 17.94 - 29.43 = -11.49 \, \text{m/s}$

Answer:

$-11.5$ m/s (rounded to 3 significant figures)