For a projectile, the square of its horizontal range is equal to 8 times the square of the maximum height it reached. What is the angle of projection of the projectile?

Step-by-step solution

The horizontal range R of a projectile is given by R = u^2 (2 ) g = 2u^2 g . The maximum height H of a projectile is given by H = u^2 ^2 2g . The problem states that R^2 = 8H^2 . Substituting the expressions for R and H : ( 2u^2 g )^2 = 8 ( u^2 ^2 2g )^2 4u^4 ^2 ^2 g^2 = 8 ( u^4 ^4 4g^2 ) 4u^4 ^2 ^2 g^2 = 2u^4 ^4 g^2 Dividing both sides by 2u^4 ^2 g^2 (assuming 0 ): 2 ^2 = ^2 ^2 = 2 = 2 = ⁻¹( 2 ) Answer: = ⁻¹( 2 )