# Aeroplane velocity problem

1. Oct 1, 2005

### brandon26

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?

2. Oct 1, 2005

### Päällikkö

You've mixed the x and y directions.
$$u_y \ne 400 \sin 88^o$$

3. Oct 1, 2005

### brandon26

Can you explain please? I thought the x direction was 400cos2 and y direction was 400sin88?

4. Oct 1, 2005

### Päällikkö

"climbing at an angle of 2 degrees" means 2 degrees from the horizontal.
Draw a diagram, and you'll see that cos is used for the x-direction.

sin gets its maximum at 90 degrees while cos reaches it at 0. ~400 m/s vertically and ~0m/s horizontally is a steep climb for an aeroplane.
Oh, and cos 2o = sin 88o

5. Oct 1, 2005

### brandon26

Yes, so the x direction is 400cos2, and y direction is 400sin88?

6. Oct 1, 2005

### Päällikkö

No.
That would mean vx = vy, which is certainly not the case here. That'd mean the elevation was 45 degrees.

For simplicity, use the same angle to express the velocities.

7. Oct 1, 2005

### brandon26

Oh right!. So Vx=400cos2 and Vy=400cos88.

8. Oct 1, 2005

### Päällikkö

Correct, although I would've used 400cos2 and 400sin2.

9. Oct 1, 2005

### brandon26

Yes, that true.