Need help with this problem in trigonometrically The image contains the following text:"37. An airplane with a speed of 97.5 m/s is climbing upward at an angle of 50.0° with respect to the horizontal. When the plane's altitude is 732 m, the pilot releases a package. (a) Calculate the distance along the ground, measured from a point directly beneath the point of release, to where the package hits the earth. (b) Relative to the ground, determine the angle of the velocity vector of the package just before impact."There are no graphs or diagrams present in the image.

Solution: Given Values, Speed (v)=97.5 m/sAngle (θ) =500Altitude (h)=732 ma) The distance along the ground is calculated that is called Range as follows: Here, The horizontal component of velocity ux=v cos 500=97.5×cos 500=62.67 m/s The vertical component of velocity uy=v sinθ =97.5 ×sin500 =74.69 m/sThe acceleration in the x-direction is zero, So it can be written as: x=R (distance)=vt+12at2x=R=uxtR=62.67 t In y-direction calculate 't' as follows: h=uyt+12gt2-732=74.69 t-12×9.8×t2By solving for t we get:t=22.03 sec Substitute t value in the Range equation: R=62.67 tR=62.67×22.03 s =1380 mThus, The distance along the ground is 1380 m b) Here, down the ground ux=vx=62.67 m/svy=u+at =74.69 m/s +(9.8 m/s ×22.03 s =-141.20 Now, The angle of velocity vector just before the impact is calculated as:Tanθ=141.20462.67θ=66.070 which is below the horizontal

Answered: *37. ssm An airplane with a speed of…