Projectile Motion: Dropping a Package

[Metamorphose]

Hi all! I just wanted to confirm whether or not the working I did for this question is wrong or right.

Here is the question:

An airplane is dropping an ad package in a remote area. The plane is moving horizontally at speed Vo and the package lands a horizontal distance L from where it was released by the plane.

Our prof. doesn't let us use memorized equations, we have to derive everything from the equation for acceleration.

[a.] Find the time it takes for the package to land.

Vx = ∫ax ∴ Vx = ax(t) + Vo.

x = ∫Vx ∴ x = 0.5ax(t)^2 + Vo(t) + xo.

The initial distance (x0) is 0 and the final distance (x) = L,

∴ L = 0.5ax(t)^2 + Vo(t)

Manipulating the equation for Vx gives ax = (Vx - Vox)/t.

Plugging this in gives L = 0.5(Vx - Vo)/t (t)^2 + Vo(t)

2L = t(Vx - Vo + 2Vo) ---> t = 2L/(Vx + Vo).

Since we can only give our answer in terms of Vo, L and g,

The final answer is t = 2L/(Vo) because the final velocity will be 0 m/s, at which point the package will have landed.


[b.] Find the altitude of the plane.

Vy = ∫ay ---> ay(t) + Voy

Y = ∫Vy ---> 0.5ay(t)^2 + Voy(t) + y0

ay = -g and y = 0,

∴ -0.5g(t)^2 + voy(2L/vox) + y0 = 0

Solving for y0 gives y0 = gL/v0

[c.] Find the velocity (vector) of the package when it lands.

So we're just going to find the velocity in the y direction as well as the x direction

Vy = ay(t) + voy.

voy = 0, and ay = -g,

therefore Vy = [-g(2L/Vox)]j

As for Vx,

Vx = [Vo]i

∴V = [Vo]i - [g(2L/Vox)]j


[d.] Find the speed of the package when it lands.

If I remember correctly, speed is just the magnitude of velocity, which would be V = (Vx^2 + Vy^2)^0.5

∴ V = [(Vo)^2 + (-g(2L/Vox))^2]^0.5

V = [Vo^2 + (4g^2(L^2))/V0x^2))^0.5


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I would greatly appreciate it if someone could read over this and tell me where I made mistakes. I know that ax in the x direction is always 0, which should alter part a into L/Vox instead of 2L/Vox.

 

[haruspex]

Sorry, but this completely wrong. Looks to me like you're not thinking about what the equations mean.
Vo is horizontal, whereas the acceleration is vertical. So it makes no sense to write x = 0.5ax(t)^2 + Vo(t) + xo.
It would also be wrong to take the final velocity as 0 because it has landed! The impact itself is not allowed for in the equation, so you have to take the final velocity as what it was just before impact.

If we ignore air resistance, the vertical and horizontal motions are quite independent. If you just think about the horizontal motion you will be able to answer [a] easily. Indeed, you seem to have done that correctly in the footnote.
Having found t, thinking only about vertical motion will allow you to solve and the vertical component of [c]. Finally, bring in the horizontal component of speed on landing to answer [c] completely.

 

[Metamorphose]

Thank you for taking the time to look over this. We are only allowed to answer in terms of vo, g and L

 

[Metamorphose]

I was actually looking over it again, and I noticed that when you are initially integrating Vx, which gives ax(t) + Vo, because ax IS essentially 0, Vx simply becomes equal to Vo. Integrating this to give x gives Vo(t) + xo. Based on the conditions given above, xo is 0 and x = L. Solving for time:

x = Vo(t) + xo

L = Vo(t)

t = L/Vo.

Does this seem to be correct?

 

[greentlc]

I'm just looking at your last post and answering this on my phone, so I'll be brief. Integrating a velocity does not give and acceleration, but a displacement. The derivative of velocity is acceleration.

 

[Steely Dan]

x = Vo(t) + xo

L = Vo(t)

t = L/Vo.

Does this seem to be correct?

Yes, that's exactly it. No need to think about the vertical direction if you can answer it by just considering the horizontal direction alone. You can then use that time to answer part (b).

 

[haruspex]

I was actually looking over it again, and I noticed that when you are initially integrating Vx, which gives ax(t) + Vo, because ax IS essentially 0, Vx simply becomes equal to Vo. Integrating this to give x gives Vo(t) + xo. Based on the conditions given above, xo is 0 and x = L. Solving for time:

x = Vo(t) + xo

L = Vo(t)

t = L/Vo.

Does this seem to be correct?

That appears to be following what I suggested and indeed gets the right answer.
I do have some problems with your description and notation.
You don't mean "integrating Vx", but rather "integrating ax to obtain Vx".
The notation Vo(t) suggests Vo as a function of t. You mean simply Vo multiplied by t, which should be written Vo.t or (in software style) Vo*t.
Can you now get and [c]?