Respuesta :
Answer:
15° and 75°
Explanation:
Given:
The range of a projectile motion is given as:
[tex]R=\frac{V_0^2sin2\theta}{g}[/tex]
Where,
[tex]V_0\to initial\ velocity\\\\\theta\to angle\ of\ projection\\\\g\to\ acceleration\ due\ to\ gravity=9.8\ m/s^2[/tex]
Maximum possible range is when [tex]\sin 2\theta=1[/tex]
So, maximum range possible is, [tex]R_{max}=\frac{V_0^2}{g}[/tex]
Now, as per question:
[tex]R=\frac{R_{max}}{2}\\\\\frac{V_0^2\sin 2\theta}{g}=\frac{V_0^2}{2g}\\\\\sin 2\theta=\frac{1}{2}\\\\\sin 2\theta=\sin 30\ or\ sin 2\theta=\sin 150\\\\2\theta=30\ or\ 2\theta=150\\\\\theta=\frac{30}{2}=15\ or\ \theta=\frac{150}{2}=75[/tex]
Therefore, the launch angles are 15 degree and 75 degree.
