Answer:
The other angle is 30 degrees.
Explanation:
The range of projectile is given by :
[tex]R=\dfrac{u^2\ \sin2\theta}{g}[/tex]
Here,
u is the speed of launch of projectile
Here, [tex]\theta=30^{\circ}[/tex]
We need to find the other launch angle when the projectile have the same range, such that,
[tex]\dfrac{u^2\ \sin(60)}{g}=\dfrac{u^2\ \sin2\alpha}{g}[/tex]
[tex]\sin(60)=\sin2\alpha[/tex]
[tex]\dfrac{\sqrt3}{2}=\sin2\alpha[/tex]
[tex]\alpha =30^{\circ}[/tex]
So, the other angle is 30 degrees. Hence, this is the required solution.