Answer:
8.66 m
Explanation:
[tex]v_{o}[/tex] = initial speed of person = 9.2 m/s
[tex]\theta[/tex] = Angle of launch = 45°
Consider the motion of the person along the vertical direction.
[tex]v_{oy}[/tex] = initial speed of person = [tex]v_{o} Sin\theta = 9.2 Sin45 = 6.51 ms^{-1}[/tex]
[tex]a_{y}[/tex] = acceleration due to gravity = [tex]- 9.8 m/s^{-2}[/tex]
[tex]t[/tex] = time of travel
[tex]y[/tex] = vertical displacement = [tex]0 m[/tex]
Using the kinematics equation that suits the above list of data, we have
[tex]y = v_{oy} t + (0.5) a_{y} t^{2} \\0 = (6.51) t + (0.5) (- 9.8) t^{2}\\t = 1.33 s[/tex]
Consider the motion of the person along the horizontal direction.
[tex]v_{ox}[/tex] = speed of person = [tex]v_{o} Cos\theta = 9.2 Cos45 = 6.51 ms^{-1}[/tex]
[tex]X[/tex] = Horizontal distance traveled
[tex]t[/tex] = time taken = [tex]1.33 s[/tex]
Since there is no acceleration along the horizontal direction, we have
[tex]X = v_{ox} t \\X = (6.51) (1.33)\\X = 8.66 m[/tex]
