A particle is projected from the ground with an initial velocity u such that the maximum height attained by it is exactly half of its horizontal range. The maximum height attained by the particle is given by :

Step-by-step solution

Let the angle of projection be . Given that the maximum height is half of the horizontal range: H = R 2 We know the formulas for maximum height and horizontal range: H = u^2 ^2 2g R = u^2 2 g = 2u^2 g Substituting these into the given condition: u^2 ^2 2g = 1 2 ( 2u^2 g ) 2 = = 2 From = 2 , we can construct a right-angled triangle to find : = 2 5 Now, substitute back into the formula for maximum height: H = u^2 2g ( 2 5 )^2 = u^2 2g ( 4 5 ) = 2u^2 5g Answer: 2u^2 5g