Answer :
Answer:
[tex]v_0 = 3.53~{\rm m/s}[/tex]
Explanation:
This is a projectile motion problem. We will first separate the motion into x- and y-components, apply the equations of kinematics separately, then we will combine them to find the initial velocity.
The initial velocity is in the x-direction, and there is no acceleration in the x-direction.
On the other hand, there no initial velocity in the y-component, so the arrow is basically in free-fall.
Applying the equations of kinematics in the x-direction gives
[tex]x - x_0 = v_{x_0} t + \frac{1}{2}a_x t^2\\63 \times 10^{-3} = v_0t + 0\\t = \frac{63\times 10^{-3}}{v_0}[/tex]
For the y-direction gives
[tex]v_y = v_{y_0} + a_y t\\v_y = 0 -9.8t\\v_y = -9.8t[/tex]
Combining both equation yields the y_component of the final velocity
[tex]v_y = -9.8(\frac{63\times 10^{-3}}{v_0}) = -\frac{0.61}{v_0}[/tex]
Since we know the angle between the x- and y-components of the final velocity, which is 180° - 2.8° = 177.2°, we can calculate the initial velocity.
[tex]\tan(\theta) = \frac{v_y}{v_x}\\\tan(177.2^\circ) = -0.0489 = \frac{v_y}{v_0} = \frac{-0.61/v_0}{v_0} = -\frac{0.61}{v_0^2}\\v_0 = 3.53~{\rm m/s}[/tex]