Answer:
Velocity (magnitude) is 98.37 m/s
Explanation:
We use the vertical component of the initial velocity, which is:
[tex]v_{0y}=v_0*sin(45)=\frac{\sqrt{2} }{2}v_0[/tex]
Using kinematics expression of vertical velocity (in y direction) for an accelerated motion (constant acceleration, which is gravity):
[tex]v_{y}=v_{0y}+a*t=\frac{\sqrt{2} }{2}v_0-9.8t[/tex]
Now we need to find [tex]v_y[/tex] as a function of [tex]v_0[/tex]. We use the horizontal velocity, which is always the same as follow:
[tex]v_x=v_0cos(45\º)=\frac{\sqrt{2} }{2}v_0=v_{t=3}*cos(30\º) \\[/tex]
We know the angle at 3 seconds:
[tex]v_y(t=3)=v_{t=3}*sin(30\º)\\v_{t=3}=\frac{v_y}{sin(30\º)}[/tex]
Substitute [tex]v_{t=3}[/tex] in [tex]v_x[/tex] and then solve for [tex]v_y[/tex]
[tex]\frac{\sqrt{2} }{2}v_0=\frac{v_y*cos(30\º) }{sin(30\º)} \\v_y=\frac{\sqrt{6} }{6}v_0[/tex]
With this expression we go back to the kinematic equation and solve it for initial speed
[tex]\frac{\sqrt{6} }{6} v_0 =\frac{\sqrt{2} }{2}v_0-29.4\\v_0(\frac{\sqrt{6}-3\sqrt{2}}{6} )=-29.4\\v_0=98.37 m/s[/tex]
