consider the motion in x-direction
[tex]v_{ox}[/tex] = initial velocity in x-direction = ?
X = horizontal distance traveled = 100 m
[tex]a_{x}[/tex] = acceleration along x-direction = 0 m/s²
t = time of travel = 4.60 sec
Using the equation
X = [tex]v_{ox}[/tex] t + (0.5) [tex]a_{x}[/tex] t²
100 = [tex]v_{ox}[/tex] (4.60)
[tex]v_{ox}[/tex] = 21.7 m/s
consider the motion along y-direction
[tex]v_{oy}[/tex] = initial velocity in y-direction = ?
Y = vertical displacement = 0 m
[tex]a_{y}[/tex] = acceleration along x-direction = - 9.8 m/s²
t = time of travel = 4.60 sec
Using the equation
Y = [tex]v_{oy}[/tex] t + (0.5) [tex]a_{y}[/tex] t²
0 = [tex]v_{oy}[/tex] (4.60) + (0.5) (- 9.8) (4.60)²
[tex]v_{oy}[/tex] = 22.54 m/s
initial velocity is given as
[tex]v_{o}[/tex] = sqrt(([tex]v_{ox}[/tex])² + ([tex]v_{oy}[/tex])²)
[tex]v_{o}[/tex] = sqrt((21.7)² + (22.54)²) = 31.3 m/s
direction: θ = tan⁻¹(22.54/21.7) = 46.12 deg
