1. The problem statement, all variables and given/known data A particle is projected with speed v at an angle α to the horizontal. Find the speed of the particle when it is at a height h. Answer: √v2 - 2gh 2. The attempt at a solution a. Since PE gained is KE lost we need to find KE and PE. KE = 0.5 * m * (v sin α)2 (the vertical component of velocity) PE = mgh 0.5m * (v sin α)2 = mgh v = √ 2gh / (sin α)2 or b. v2 = u2 + 2as Upward direction is positive. So we have: u = v sin α a = -g s = h v = √(v sin α)2 -2gh But both of them do not fit the answer. Thanks in advance for any help.