To solve this problem we will apply the linear motion kinematic equations. We will find the two components of velocity and finally by geometric and vector relations we will find both the angle and the magnitude of the vector. In the case of horizontal speed we have to
[tex]v_x = \frac{x}{t}[/tex]
[tex]v_x = \frac{67}{4.5}[/tex]
[tex]v_x = 14.89m/s[/tex]
The vertical component of velocity is
[tex]-h = v_y t -\frac{1}{2} gt^2[/tex]
Here,
h = Height
g = Gravitational acceleration
t = Time
[tex]v_y[/tex] = Vertical component of velocity
[tex]-1.23 = v_y(4.5)-\frac{1}{2} (9.8)(4.5)^2[/tex]
[tex]-1.23= 4.5v_y - 99.225[/tex]
[tex]v_y = 21.77m/s[/tex]
The direction of the velocity will be given by the tangent of the components, then
[tex]tan\theta = \frac{v_y}{v_x}[/tex]
[tex]\theta = tan^{-1} (\frac{21.77}{14.89})[/tex]
[tex]\theta = 55.59\°[/tex]
The magnitude is given vectorially as,
[tex]|V| = \sqrt{v_x^2+v_y^2}[/tex]
[tex]|V| = \sqrt{14.89^2 +21.77^2}[/tex]
[tex]|V| = 26.37m/s[/tex]
Therefore the angle is 55.59° and the velocity is 26.37m/s
