so i am having issues with finding the distance for this question. here is the question and what i tried to do to solve it. 1. The problem statement, all variables and given/known data An airplane with a speed of 17.6 m/s is climbing upward at an angle of 42° counterclockwise from the positive x axis. When the plane's altitude is 840 m the pilot releases a package. Calculate the distance along the ground, measured from a point directly beneath the point of release, to where the package hits the earth. known: v (initial) = 17.6 m/s direction 42 degrees height (horizontal) = -840 m a in the y direction: -9.8 m/s^2 a in the x direction: 0 m/s^2 unknown: v initial in the y or x direction time distance (vertical) 2. Relevant equations d=v(initial)*t+1/2*a*t^2 in the y direction and d = (avg velocity)(time) in the x direction 3. The attempt at a solution the question is asking what is the distance from the point of the release of the package to when it hits the ground. i calculated the time it would take for the package to reach the ground and used v initial in the y-direction to be 0 since that is the initial velocity in the y direction when at the max height of the projectile (-840m for the package). i plugged into the equation d=v(initial)*t+1/2*a*t^2 --> -840 = 0*t+1/2*-9.8*t^2 = 13.1 s. THEN i found that v intial in the x direction would be 13.1 m/s using trigonometry (13.1=cos42*17.6m/s). By plugging into d=vt, i get 13.1 m/s * 13/1 s = about 171 m. apparently this is the wrong answer. can anyone help me with this? thanks!