# Kinematics in two dimensions

#### kcalhoun

1. The problem statement, all variables and given/known data

An airplane with a speed of 59.6 m/s is climbing upward at an angle of 40° counterclockwise from the positive x axis. When the plane's altitude is 600 m the pilot releases a package.
(a) Calculate the distance along the ground, measured from a point directly beneath the point of release, to where the package hits the earth.

(b) Determine the angle of the velocity vector of the package just before impact.

2. Relevant equations

Eight kinematics equations

3. The attempt at a solution

completely stuck.. i don't even know where to begin
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

#### The Bob

Welcome to PF kcalhoun!

This question requires you to split the problem into velocity components and then use your constant acceleration formulae. Also think about what the package does once it is released. So to start with, can you tell us what the speed is of the plane in the horizontal (positive x) and vertical directions (positive y)?

The Bob

#### kcalhoun

for the horizontal it would be 59.6cos40
and the vertical would be 59.6sin40

#### The Bob

Nice. So, if we can find out how long the package takes to fall to the ground, then it would seem logical that we can find out how far it travels due to time. So what do we now need to consider?

The Bob

#### kcalhoun

solve for t in one of the constant acceleration equations?

#### The Bob

Yes, in short. However, remember to consider the conditions the package is under once released. What is it's inital vertical velocity? etc. The key is what is positive and what is negative.

The Bob

#### kcalhoun

ok so using the equation vy=v0y+ayt i get.. 0=59.6sin40+(-9.80)t and t = 3.909?

#### The Bob

That wasn't the one I was thinking of. What you've worked out is the time it takes for the package to come to rest whilst in mid-air.

The Bob

#### kcalhoun

I'm not sure what you are looking for then...

#### The Bob

Okay. Once the package is released, it continues to travel upwards as it still has a vertical component. Gravity takes effect and slows it down such that its velocity after 3.909s is 0ms-1. So, you can either start over with one equation that will do it all for you (e.g. to get total time) or you can stick with this and you can now work out how far it has travelled vertical in addition to the 600m and thus the total time for it to be released and fall to the ground.

Do you understand why what you've worked out is the time for it to be at v = 0 in mid-air?

The Bob

#### kcalhoun

not really no but, was this the equation that you were thinking of... y=v0y+1/2ayt^2?

#### The Bob

That was the one I was thinking of but I'd rather you understood why. The plane ascends and the package is released. The package has the same vertical component as the plane and it is immediately constant, e.g. no external force, after released. However, gravity now acts on the package. (At this point I should note that I will assume negative y is actually positive and positive y is negative but it doesn't really matter that much.) So the package is travelling at -59.6sin(40). This is our initial velocity. We also know at what height above the ground the package was released (600m) and we know that a = g = 9.8ms-2. Does that help you see what I'm getting at? Also, the reason t = 3.909s is write for what I said is because the formula automatically works out the next time something happens (in this case when v = 0).

The Bob

#### kcalhoun

ok yeah that makes sense so now when i solve for t i get 114.631 then i need to find delta x to solve part a?

#### The Bob

The equation is:
s = u*t + 0.5*a*t^2
where s is the displacement, u is the intial velocity, t is time and a is the constant acceleration. Your equation in post 11 had a missing t, my bad... sorry. I'm not doing very well here.

The Bob

#### kcalhoun

right i used that equation to solve for t and i got 600=59.6sin40t+1/2(-9.8)t^2 which yields 114.631 right? then where do i go from there?

#### The Bob

Can you put your working out up please? What you have will yield an imaginary number.

The Bob

#### kcalhoun

(600-59.6sin40)/(1/2*9.8)=114.631 does that look right?

#### The Bob

Where have your 't's gone????? You can't solve for t without there being any in the equation. This is a quadratic equation so you're need the quadratic formula.

The Bob

#### kcalhoun

what i did was take the original equation y=v0yt+1/2ayt^2 and divided by t to get y=v0y+1/2ayt and the i solved for t to get t=(y-v0y)/(1/2ay) and then i plugged in the numbers.. t=(600-59.6sin40)/(1/2*9.8) so t=114.631 is that not right?

#### The Bob

It's a good idea but it will not work here because 't' could be positive or negative. You can only divide if you know it is always positive or always negative. So now try it as a quadratic.

The Bob

P.S. Sometimes it's worth thinking logically. It only takes an object, initially at rest, just over 11 seconds to fall 600 metres so 114.631 seconds should stand out as not being right.

### The Physics Forums Way

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving