Answer:u=42.29 m/s
Explanation:
Given
Horizontal distance=167 m
launch angle[tex]=33.1^{\circ}[/tex]
Let u be the initial speed of ball
Range[tex]=\frac{u^2\sin 2\theta }{g}[/tex]
[tex]167=\frac{u^2\sin (66.2)}{9.8}[/tex]
[tex]u^2=1788.71[/tex]
[tex]u=\sqrt{1788.71}[/tex]
[tex]u=42.29 m/s[/tex]