Solving Kinematic Q: Plane Releasing Package, Angle of Velocity Vector

[Kalu]

I've been trying this question, but I doubt the first answer...

An airplane with speed 97.5m/s is climbing upward at an angle of 50.0degrees with respect to the horizontal. When the plane's altitude is 732m, the pilot releases a package.
(a) Calculate the distance along the ground, measured from a point directly beneath the point of release, to where package hits the ground.
(b) Relative to the ground, determine the angle of velocity vector of package
just before impact.

For (a): I started used this equation Y = (Vf^2- Vi^2)/2a; plugged in the given values (97.5)^2/(2*9.8) = 485m. Is this correct. How do I start Q(B)?

I appreciate the help.

 

[arildno]

Why should you be able to use that equation in a)?
You don't know the the final speed..
In addition, your answer is meaningless; you were asked to find the distance traveled along the ground (measured from...).
Your answer contains no such information.
To give you a few hints:
1. Determine the initial vertical component of the package's velocity.
2. Calculate the time it will take to reach ground level.
3. Use that time to determine the horizontal distance traveled.

 

[M98Ranger]

It looks like you are on the right track for part (a) of the question. However, instead of using the equation Y = (Vf^2- Vi^2)/2a, you should use the equation Y = Vi*t + 1/2*a*t^2, where Vi is the initial velocity, a is the acceleration (in this case, due to gravity), and t is the time. Since the plane is moving at an angle, the initial velocity in the vertical direction would be Vi = 97.5m/s*sin(50.0degrees) = 75.0m/s. The time can be found using the equation Y = Vi*t + 1/2*a*t^2 and solving for t. Once you have t, you can plug it into the equation X = V*t to find the distance along the ground.

For part (b), you can use the fact that the velocity in the horizontal direction remains constant at 97.5m/s, while the velocity in the vertical direction decreases due to the acceleration of gravity. You can use the equation Vf = Vi + a*t to find the final velocity in the vertical direction just before impact. Then, you can use the tangent function to find the angle of the velocity vector relative to the ground. It would be tan^-1(Vf/Vhf), where Vhf is the horizontal component of the velocity.