At what height h above the ground does the projectile have a speed of 0.5v?
K_1 + U_1 = K_2 + U_2
K = 1/2mv^2
U = mgh
The Attempt at a Solution
I made the point where the speed is .5v = point 1, and the maximum height = point 2.
K_1 = 1/2m(v_1x^2 + v_1y^2)
U_1 = mgh
K_2 = 1/2m(v_2x^2 + v_2y^2)
U_2 = mg(v^2/2g) = (mv^2)/2 <--maximum height, I got this from the previous part.
Then it's just plug and chug into the eqn K_1 + U_1 = K_2 + U_2 but I came out with the wrong answer which is (-v^2/8g). Oh I took out the v_1x^2 and the v_2x^2 because the x component remains constant throughout and I also took v_2y^2 because it's 0 at maximum height.
I must have done something wrong somewhere but I have no idea (probably my setup) so pls help?