Answer:d)Both
Explanation:
Given
Projected velocity (u)=5 m/s
First Projectile angle[tex](\theta _1)=10^{\circ}[/tex]
Second Projectile angle[tex](\theta _2)=80^{\circ}[/tex]
Range is given by [tex]R=\frac{u^2sin2\theta }{g}[/tex]
[tex]R_1=\frac{5^2sin20}{9.81}=0.871 m[/tex]
[tex]R_2=\frac{5^2sin160}{9.81}=0.871 m[/tex]
Range of both projectiles are same as they launched at complementary angles.