What was the plane's speed when the bottle was released?

[onigiri]

Homework Statement

The movie 'The Gods Must Be Crazy' begins with a pilot dropping a bottle out of an airplane. It is recovered by a surprised native below, who thinks it is a message from the gods. If the plane from which the bottle was dropped was flying at a height of 500m and the bottle lands 400m horizontally from the initial dropping point, how fast was the plane flying when the bottle was released?

Homework Equations

I know this isn't an equation but I think it might be relevant to know the accelaration rate... I'm to use g=10m/s/s for this question.

The Attempt at a Solution

So I've drawn myself a picture on my notebook with a plane traveling and this guy dropping a bottle... and I've made different frames, showing that the plane is traveling little by little.
Well. To list the knowns and unknowns, I have this:
distance from ground to plane: 500m
distance between dropping point and landing point (horizontally): 400m
speed the plane was traveling at: UNKNOWN

We just started this section... and I'm struggling. I don't want to feel like I should never have taken IB high physics... at this rate, I SO will. Please help me understand this!

 

[Gokul43201]

You need to know and understand the equations of motion. Can you write down all the equations?

 

[Chi Meson]

Have you done any projectile motion problems before? You know, cannonball shot from a cliff, golf ball hit down the fairway, soccer ball kicked up on a roof?

 

[onigiri]

First Post:
I've never had physics before this year- it's new to me, and since I'm an IB student, I have nothing memorized. I don't even know what equations of motions mean... I mean like, what kind of motion...?

Second Post:
No, we've discussed things and tried to understand concepts, but we've never actually solved any problems. We haven't had any problems with acutuall numbers or anything either- just stuff like 'so the ball is rolling off at a constant speed and it's a horizontal force. Since gravity is a vertical force, neither has any effect on each other.' So yes... I really am kind of starting from scratch.

 

[Chi Meson]

You have two types of motion, vertical and horizontal. Both happen at the same time, but they are separate. In the following equations, the subscript x refers to the horizontal components and y for the vertical components.
The horizontal motion happens at constant velocity, and has one equation takes care of that:
$$d _{x} = v_{x} t$$

The vertical motion happens while undergoing uniform acceleration (due to gravity). In all of the following use
$$a = -9.807 \frac {m}{s\ ^{2} }\$$

Depending on what you are given, you will generally use one of the following four equations to find an unknown:
$$v_{fy} = v_{oy}+at$$

$$v^{2}_{fy} = v^{2}_{oy} +2ad_{y}$$

$$d_{y}=v_{oy}t + \frac {1}{2}\ at^{2}$$

$$d_{y}=\frac {v_{oy}+v_{fy}}{2}\ t$$

Set up two columns, one for the horizontal information, and one for the vertical. Using angles given (usually the initial angle of velocity) find the horizontal and vertical components of the initial velocity

$$v_{x}=v_{o}cos \theta$$
$$v_{oy}=v_{o}sin \theta$$
(here, theta is the angle that the initial velocity makes with the horizontal surface)

Examine the problem for as many of the values that go with these equations. remember, there are three variables for horizontal motion (constant velocity), but five variables for vertical motion (uniform acceleration).

Next, determine which unknown variable will answer (or help to answer) the question as stated in the problem.

In your two columns, the one factor that is the same for both is the time, t. When you have solved for t in one column, then you can drag that value to the other column. You will be told (including inferences and assumptions) either two of the three values for the horizontal information, or three of the five variables fore the vertical information.

Often, you need to find the t in one column, drag it to the other column and use that t to solve for the wanted unknown variable.

 

[onigiri]

so... v=u+at

right? velocity=initial velocity plus accelaration times time?

and v^2=u^2+2as