What is the angle of the velocity vector of a released package?

[mac227]

Homework Statement

An airplane with a speed of 96.8 m/s is climbing upward at an angle of 74.0 ° with respect to the horizontal. When the plane's altitude is 906 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.

Homework Equations

The Attempt at a Solution

I solved for part a. The distance x is 695.78m

I now need to determine the angle of the velocity vector of the package just before impact. I'm having trouble with this part, here is what I have so far:
v=96.8
v_oy=0
a_y=9.8
v_y=?
t=
y=906
theta=?


v²_y = v_oy + 2(a_y)y

v_y=133.24 m/s

theta=sin^-1 (v_y/v)

v= square root of vx² + vy² vox=vx vx=26.68

v= 165.9

so

theta = 53.43 this answer is wrong I am not sure where I went wrong

 

[verty]

I used v = ds/dt, tan(theta) = v_y / v_x. I didn't realize that the velocity would still be 96.8 which makes it a little simpler.

Looking at what you have done, why does V0y = 0?

 

[mac227]

I used v = ds/dt, tan(theta) = v_y / v_x. I didn't realize that the velocity would still be 96.8 which makes it a little simpler.

Looking at what you have done, why does V0y = 0?

because i was calculating the velocity in the y direction of the package being dropped. WHen the plane reaches 906m it drops a package. So that packages initial velocity is 0. voy=0


So to get theta you used tan(theta) = v_y / v_x ?

did you get 26.68 for vx and how would you go about calculating vy? is this where you use v = ds/dt? what does ds stand for? dt is distance*time correct?

 

[verty]

When the plane reaches 906m it drops a package. So that packages initial velocity is 0.

It is 0 relative to the plane, but the plane was not stationary at the time.

Velocity is the derivative of displacement, is what v = ds/dt means. Using calculus, I got a formula for velocity versus time. You wouldn't need to do this though, I only suggested it because it is a more conceptual method (using calculus).

Try to solve it now (using any method) with the correct initial velocity.

 

[verty]

Oh sorry, the velocity varies in magnitude after the package is dropped. For some reason I thought it did not.

So, how to solve this? One needs two of v_x, v_y, v. V_x is easy to calculate, v_y is easier to calculate than v I think. I would try to do that.

 

[mac227]

solve for the vy going up or down?

v_x is the same as v_ox because a_x = 0 correct?

v_x = 26.68

v_y= 0 at peak... calculate v_y coming down??