Answer:
True, [tex]x_r[/tex] = 1353 m
Explanation:
Given:
[tex]v_o=125\ m/s[/tex]
[tex]\theta=30^o[/tex]
[tex]g=10\ m/s^2[/tex]
[tex]\boxed{range(x_r)=\frac{v_o^2}{g}sin(2\theta) }[/tex]
[tex]x_r=\frac{125^2}{10} sin(2\cdot30^o)[/tex]
[tex]=1562.5\times sin(60^o)[/tex]
[tex]=1562.5\times \frac{1}{2} \sqrt{3}[/tex]
[tex]=1353\ m[/tex]
Therefore, the statement is true
