A ball is thrown straight upward and returns to the thrower's hand after 2.50 s in the air. A second ball is thrown at an angle of 25.0° with the horizontal. At what speed must the second ball be thrown so that it reaches the same height as the one thrown vertically? m/s okay, I am not sure what I am doing wrong here: to find deltaY it would be deltay = -1/2(g)(t)^2 = deltay -1/2(-9.8)(2.50)^2 deltay = 30.63 now i know that deltaX = 30.63 because that is the same distance the first ball is. V_0y = Vosin25 and V_0x = Vocos25 when setting up the eqaution i should get 30.63 = (V_0^2)/(9.8) x sin2(25) 30.63/sin2(25) = Vo^2/9.8 which then become Vo^2 = 39.98 Vo = 6.32 which is incorrect. What mistake or what steps should I be following.