Hi all! I just wanted to confirm whether or not the working I did for this question is wrong or right.(adsbygoogle = window.adsbygoogle || []).push({});

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.

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# Homework Help: Projectile Motion: Dropping a Package

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