Answer:
22.1 m
Explanation:
[tex]v_{o}[/tex] = initial speed of ball = 14.3 m/s
[tex]\theta[/tex] = Angle of launch = 27°
Consider the motion of the ball along the vertical direction.
[tex]v_{oy}[/tex] = initial speed of ball = [tex]v_{o} Sin\theta = 14.3 Sin27 = 6.5 ms^{-1}[/tex]
[tex]a_{y}[/tex] = acceleration due to gravity = - 9.8 ms⁻²
[tex]t[/tex] = time of travel
[tex]y[/tex] = vertical displacement = - 3.50 m
Using the kinematics equation that suits the above list of data, we have
[tex]y = v_{oy} t + (0.5) a_{y} t^{2} \\- 3.50 = (6.5) t + (0.5) (- 9.8) t^{2}\\- 3.50 = (6.5) t - 4.9 t^{2}\\t = 1.74 s[/tex]
Consider the motion of the ball along the horizontal direction.
[tex]v_{ox}[/tex] = initial speed of ball = [tex]v_{o} Cos\theta = 14.3 Cos27 = 12.7 ms^{-1}[/tex]
[tex]X[/tex] = Horizontal distance traveled
[tex]t[/tex] = time taken = 1.74 s
Since there is no acceleration along the horizontal direction, we have
[tex]X = v_{ox} t\\X = (12.7)(1.74)\\X = 22.1 m[/tex]
