Answer:
74.28 m/s
Explanation:
In this case, it is a zero launch angle in that the object is launched horizontally at a height H.
The formula to apply is:
[tex]R=v_o*\sqrt{\frac{2H}{g} }[/tex]
where
R= range, 450 m
H= height of building ,180 m
g= 9.81
vo= ? initial velocity
Applying the values to the equation;
[tex]450 = v_o*\sqrt{\frac{2*180}{9.81} } \\\\450=v_o*\sqrt{36.70} \\\\450=v_o*6.06\\\\450/6.06 = v_o\\\\74.28 m/s=v_o[/tex]
