For two projectiles to have the same horizontal range R with the same initial speed u, their angles of projection must be complementary, i.e., θ and 90∘−θ.
The times of flight for these two angles are given by:
t1=g2usinθ
t2=g2ucosθ
Multiplying t1 and t2, we get:
t1t2=(g2usinθ)(g2ucosθ)=g2(gu2(2sinθcosθ))
Since the horizontal range is R=gu2sin2θ, we can write:
t1t2=g2R
Given t1=5 s, t2=10 s, and g=10 m/s2, substituting these values:
5×10=102R
50=102R
2R=500⇒R=250 m
Answer: 250