An aeroplane is climbing at an angle of 2 degrees while maintaining a speed of 400ms^-1. A package is released and travels a horizontal distance of 2500m before hitting the ground. The initial velocity of the package is the same as the initial velocity as the plane, find the height of the aeroplane above the ground at the moment the package was released. I did some calculations but my answer is apparantly wrong. Considering motion in the vertical plane. a= 9.8 s=? u=400sin88 t=? To find t i used the horizontal plane and got 6.25 sec. therefore s = ut + 05.at^2 (400sin88 x 6.25) + 0.5x9,8x6.25^2 =2700M Where did I go wrong?