Answer:
The actual angle is 30°
Explanation:
[tex]v_y(t)=vo*sin(A)-g*t[/tex]
the velocity is Zero when the projectile reach in the maximum altitude:
[tex]0=vo-gt\\t=\frac{vo}{g}[/tex]
When the time is vo/g the projectile are in the middle of the range.
[tex]d_x(t)=vo*cos(A)*t\\[/tex]
R=Range
[tex]R=d_x(t=2*\frac{vo}{g})[/tex]
[tex]R=vo*cos(A)*2\frac{vo}{g} \\\\R=\frac{(vo)^{2}*2* sin(A)cos(A)}{g} \\\\R=\frac{(vo)^{2} sin(2A)}{g}[/tex]
**sin(2A)=2sin(A)cos(A)
Let B the actual angle of projectile
[tex]\frac{vo^{2} }{g} =(\frac{2}{\sqrt{3} }) \frac{vo^{2} *sin(2B)}{g}\\\\1= \frac{2 }{\sqrt{3}} *sin(2B)\\\\sin(2B)=\frac{\sqrt{3}}{2}\\\\[/tex]
2B=60°
B=30°
