Motion Problem Using F=Ma and s=ut+1/2xt^2

[jakan]

Firstly, appologies if this is in the wrong place or wrote wrong, total noobie here.

Ive been given some examples of math questions based on aviation, unfortunately this one didn't come with the correct answer or any example working and it seems a little more tricky than the others so I am struggling with where to start and knowing if I am getting decent answers. (Its for an oral examination with a maths element, don't worry the scenaria and numbers I get will be completely unknown to me):

A aircraft is conducting a low-level training supply drop run with the intention of dropping a 10kg package on a target location. The aircraft is at a constant speed of 150ms^-1 at an altitute of 200m.

At package release, the package is subject to a constant acceleration in a downward vertical direction of 4ms^-2 and in the horizontal direction of -3ms^-2 (ie against direction of travel).

At what horizontal distance from target should the package be released and what will the impact force be?

Equations given are F=ma and s=ut+1/2xt^2

Any help or pointing in right direction will be appreciated, other examples used other motion equations I had no issue with.

 

[IanCg]

Firstly, appologies if this is in the wrong place or wrote wrong, total noobie here.

Ive been given some examples of math questions based on aviation, unfortunately this one didn't come with the correct answer or any example working and it seems a little more tricky than the others so I am struggling with where to start and knowing if I am getting decent answers. (Its for an oral examination with a maths element, don't worry the scenaria and numbers I get will be completely unknown to me):

A aircraft is conducting a low-level training supply drop run with the intention of dropping a 10kg package on a target location. The aircraft is at a constant speed of 150ms^-1 at an altitute of 200m.

At package release, the package is subject to a constant acceleration in a downward vertical direction of 4ms^-2 and in the horizontal direction of -3ms^-2 (ie against direction of travel).

At what horizontal distance from target should the package be released and what will the impact force be?

Equations given are F=ma and s=ut+1/2xt^2

Any help or pointing in right direction will be appreciated, other examples used other motion equations I had no issue with.

Your equation s=ut+1/2xt^2 does not seem to be correct it is usually written s=ut+1/2at^2 where a is the acceleration.

you find t by looking at the motion vertically and then s using the value of t you have found in the horizontal equation

 

[HallsofIvy]

You don't really need "F= ma" since you are not given or asked about the force. You are given the vertical and horizontal accelerations so you need s= ut+ (1/2)at^2. The package has a downward acceleration of 4 m/s^2 and must fall 200 m. u is the initial speed but there was no initial downward speed so u= 0. So for the vertical motion (1/2)(4)t^2= 2t^2= 200. Solve that for t. (Notice that the acceleration is not negative because I am taking "downward" to be positive.)

You are told that the horizontal acceleration is -3 m/s^2. The initial horizontal speed is 150 m/s so the horizontal distance is given by d= 150t- 3t^2. Use the t determined above to find the distance the package will move horizontally while falling.

 

[Dethrone]

You don't really need "F= ma" since you are not given or asked about the force. You are given the vertical and horizontal accelerations so you need s= ut+ (1/2)at^2. The package has a downward acceleration of 4 m/s^2 and must fall 200 m. u is the initial speed but there was no initial downward speed so u= 0. So for the vertical motion (1/2)(4)t^2= 2t^2= 200. Solve that for t. (Notice that the acceleration is not negative because I am taking "downward" to be positive.)

You are told that the horizontal acceleration is -3 m/s^2. The initial horizontal speed is 150 m/s so the horizontal distance is given by d= 150t- 3t^2. Use the t determined above to find the distance the package will move horizontally while falling.

Well, to be fair, the OP did ask about the impact force in the latter part of their sentence.

 

[HallsofIvy]

Ah, thanks.