Another kinematics question?

Question

An airplane with a speed of 97.5 m/s is climbing upward at an angle of 50 degrees to the horizontal. When the plane's altitudeis 732 m, the pilot releases a package. Relative to the ground, determine the angle of the velocity vector of the package just before impact.

Answer 1

First, figure the speed in the horizontal direction

Vx= Cos (theta) * 97.5 m/sec

Second, figure the Velocity in the Y direction at impact.

I'm going to cheat a little and use Potential and Kinetic Energy 'cause it's easier.

First find Vy at release:

Vyr = Sin (theta) * 97.5 m/sec

Kinetic Energy (KEr)= 1/2 mVyr^2

Note, this is up direction, but it will then go back down at the same speed

Kinetic Energy at the ground= PE + KEr= 1/2 m(Vyg)^2
where Vyg= Velocity in the y direction at the ground.

1/2mVyg^2=mg*732m + 1/2 mVyr^2

therfore,

Vyg= squroot(2g*732m+Vyr^2)

Vyg= squroot [2g*732m+ (Sin (theta) * 97.5 m/sec)^2]

You now have your two velocities vectors at the ground Vx and Vyg. Plugnchug and you have your answer.