Question

Derive an expression for horizontal range of projectile

Answer 1

Here, u_x = ucosθ

and u_y = usinθ

At maximum height, v_y = 0

⇒ u sinθ - gt = 0

⇒ \(t =\frac{usin\theta}{g}\)

So, time of flight (T) = 2t = \(\frac{2u\,sin\theta}{g}\)

and horizontal range = u cosθ × T = \(\frac{u^2\,sin\theta\,cos\theta}{g}=\frac{u^2\,sin2\theta}{g}\)