1. The problem statement, all variables and given/known data A helicopter is rising vetically at a uniform velocity of 14.7 m/s. When it is 196 m from the ground, a ball is projectted from it with a horizontal velocity of 8.5 m/s with respect to the helicopter. Calculate: a) When the ball will reach the ground b) Where it will hit the ground c) What its velocity will be when it hits the ground Answers are: 8.0 s, 68 m, 64 m/s 82 degrees below horizontal 2. Relevant equations c² = root of a² + b² d = v1 t + 1/2 a (t)² 3. The attempt at a solution So i drew out the path of the projected ball and used Pythagorean theorem to solve the 2 vectors of 14.7 [N] and 8.5 [E] and i get 16.9 m/s [N 60 E] I used d = v1 t + 1/2 a (t)² to get distance However, my answer comes out incorrect (by a lot) so i used: V2 ^ 2 - V1 ^2 = 2 a d 285 / 2(-9.8m/s^2) = d d = 15 m However, That is not the correct answer. Looking for someone to point me in the right direction, thanks.