Kinematics Plane and Rocket Problem

[majorazero]

Homework Statement

Some plane is flying at an altitude of 12.0 km above sea level and is flying in a straight line at 850km/h. The plane then drops a rocket. After the drop the plane maintains the same altitude and speed and continues to fly in a straight line. The rocket falls for a brief time, after which its motor turns on and the combined effects of thrust and gravity gives the rocket a constant acceleration of 3.00g directed at an angle of 30 degrees above the horizontal. For reasons of safety, the rocket should at least be 1.00 kn in front of the plane when it climbs through the airline's altitude. Determine the minimum time that the rocket must fall before its engine starts. You can ignore air resistance.

Homework Equations

I used the basic three kinematics equation:
v = v+at
v^2 = v^2 + 2ax
x = vt + .5at^2
some trig equations too i guess

The Attempt at a Solution

I figured that the rocket experiences free fall and should have the vertical component of -gt. Then I thought that the horizontal component should be the equivalent of the plane so it would be 860 km/h or 3060 m/s. Since its moving relative to the plane I kind of just assumed its horizontal velocity is 0 relative to the plane.

Afterward, I derived the horizontal acceleration of the rocket which would be 25.46 meters squared and figured out how long it would take for it to reach 1000 m which was 8.86 seconds. I then figured that if one of its horizontal position is 1000 I can use trig to figure out its vertical, which using tan30 = x/1000 i got 577.35 meters.

I then punched it in x= vt + .5at^2 which looked like 577.35 = -gt(8.86) + .5(29.4*sin30)(8.86^2), either way I got like -0.00435 seconds which is ridiculous. Help :(

 

[tomwilliam]

I'm no expert, but if you have found out how many metres the rocket needs to fall before it can start its thrust, and the question is asking for how long in seconds the rocket needs to freefall, the calculation is fairly simple, using g.

 

[majorazero]

Agreed, and I've tried that. Unfortunately, the computed value doesn't work.