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## Homework Statement

During a drop, a bomber decends with velocity v, directed at angle 37 degrees below horizontal. The cargo is released at an altitude of 2450 m and reaches intended dropzone with displacement of 3850 m

What is the speed of bomber when it releases its cargo

## Homework Equations

[itex]

V_{0x} + a_x t = V_x ----

\Delta x = (1/2)(V_{0x} + V_x) t ----

\Delta x = V_{0x}t + (1/2)a_x t^2 -----

\Delta x = V_x t - (1/2) a_x t^2

[/itex]

## The Attempt at a Solution

So I begin with a triangle.

Hypotenus = 3850 m. angle opposite hypotenus is 90 degrees. I find the top angle to be 53 degrees, because the plane is coming down at angle 37 degrees. That leaves angle at the drop to be 37 degrees.

Does everything seem right so far?

The length of the 37 degrees angle is 2450 m - x, but I find it using law of sines.

3850 m / sin(90) = x / sin(37)

I find it to be 2316 m, and using pythagoras I get 3075 m to be the length for angle 53 degrees.

Now, I'm being asked to find v, speed at which plane was going as cargo is dropped.

[itex] V = \sqrt {V_{0x}^2 + V_{0y}^2 } [/itex]

Now this is my second attempt at this problem and I'm wondering why this method doesnt make sense. I will also post my first method and see if its okay with y'all, I think it is but it just seems like sooo much work.

Since [itex] V_{0x} = V_x [/itex], I use [itex] \Delta x = (1/2) (V_{0x} + V_x) t [/itex] and get

[itex] V_{0x} = \Delta x /t [/itex]

For [itex] V_{0y} = xsin(37) [/itex]

I don't see anything wrong with the statements I made so far. If anyone does please let me know here.

Anyways, [itex] V_{0x} = xcos(37) = \Delta x /t [/itex]

so [itex] x = \Delta x / t cos(37) [/itex]

Now plugging x into [itex] V_{0y} [/itex]

[itex] \Delta x tan(37) / t [/itex] this = [itex] V_{0y} [/itex]

Now I want to use equation [itex] \Delta y = V_{0y} t + (1/2) a_y t^2 [/itex]

[itex] \Delta y = \Delta x tan(37) + (1/2) a_y t^2 [/itex]

Notice how I got rid of the t in the original motion equation. Then I solved:

[itex] \sqrt {2(\Delta y - \Delta x tan(37))/a_y} = t [/itex]

But the value that I get for t using this method just doesn't make sense. It's under 1, and no way can the cargo be dropped off in under 1 second.

I'm about to make a post in the comments with my other method which was much uglier, but I think would provide a correct answer... , but can anyone explain to me why what I did here doesn't make sense?