Respuesta :
Answer:
C. At the highest point of flight.
Explanation:
Given data;
[tex]u_{x} = 4 m/s ,u_{y} = 0, h = 60 m[/tex]
Time of flight,
by third equation of motion ,
[tex]h=\frac{1}{2} gt^{2}[/tex]
[tex]t=\sqrt{\frac{2*60}{9.8} }[/tex]
[tex]t = 3.5 Sec[/tex]
so the range ,
by horizontal component of velocity,([tex]u_{x}[/tex])
R = 4X3.5
R = 14 m
Horizontal component of velocity remain constant in the entire motion.