Solution:

Given the relationship \( R = 5T^2 \), and using the formulas for range \( R \) and time of flight \( T \):

1. **Range formula:**
R = u cos(θ) · T

2. **Time of flight formula:**
T = (2u sin(θ)) / g

Substituting T into the range equation:

R = u cos(θ) · (2u sin(θ) / g)

This gives:

R = (2u² sin(θ) cos(θ)) / g

Setting this equal to \( 5T^2 \):

(2u² sin(θ) cos(θ)) / g = 5((2u sin(θ)) / g)²

Simplifying:

(2u² sin(θ) cos(θ)) / g = (20u² sin²(θ)) / g²

Cancelling \( u² / g \):

2 cos(θ) = 20 sin(θ) / g

Given g = 10 m/s²:

2 cos(θ) = 2 sin(θ)

Thus:

cos(θ) = sin(θ)

This implies:

tan(θ) = 1 → θ = 45°

The angle to the horizontal should be  45°