Some particle is given an initial velocity, u at an angle θ to the horizontal. I'm asked to find (as a function of θ): 1. the range of the particle, X, 2. the maximum altitude reached, Y and the time taken to reach maximum altitude, T. First I resolved u into components: uy = usin θ, ux = ucos θ. For 1, I said x = ucos θ.t (since the horizontal motion is unaccelerated, no air resistance). To work out t, I used s = ut + (1/2)at^2 for the vertical motion, setting s = 0. For t != 0, I got t = (2usin θ)/g => X = (ucos θ.2usin θ)/g = (2u^2.sin θcos θ)/g = (u^2.sin 2θ)/g For 2, I used the fact that v = 0 when the particle reaches its maximum height and the equation v^2 = u^2 + 2as. => Y = (1/2g)(uy)^2 = (1/2g)(usin θ)^2 For 3, I again used v = 0 at maximum height, but used v = u + at => T = (usin θ)/g Then, I'm asked to work out the projectile's velocity as a function of time. v(t) = [(vx)^2 + (vy)^2]^1/2 vx = ucos θ Using v = u + at, vy = usin θ - gt v(t) = [(ucos θ)^2 + (usin θ - gt)^2]^1/2 v(t) = [(ucos θ)^2 + ((usin θ)^2 - 2gtusin θ + (gt)^2]^1/2 v(t) = [u^2.(cos^2 θ + sin^2 θ) - 2gtusin θ + (gt)^2]^1/2 v(t) = [u^2 - 2gtusin θ + (gt)^2]^1/2 Is this correct?