What is the height of the aeroplane at the moment the package was released?

[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 apparently 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?

 

[Päällikkö]

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

 

[brandon26]

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

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

 

[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 2^o = sin 88^o

 

[brandon26]

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

 

[Päällikkö]

No.
That would mean v_x = v_y, 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.

 

[brandon26]

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

 

[Päällikkö]

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

 

[brandon26]

Yes, that true.