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3d projectile motion

by thegame
Tags: motion, projectile
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thegame
#1
Sep20-03, 05:17 PM
P: 32
You are driving in a car moving at 52 km/h [E]. YOu thrown an empty pop can out the window at 22km/h [N 30 degrees W] at an angle of 15 degrees above the horizontal. The vertical displacement is 0. Where will the pop can land in relation to the starting point?
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FZ+
#2
Sep20-03, 06:28 PM
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Is that 22Km/h relative to your car, or relative to the ground?
thegame
#3
Sep20-03, 09:21 PM
P: 32
the car is moving on the road. Therefore, 22 km/h is the velocity of the car relative to the ground.

Why do you ask? I don't think it really matters. ;)

HallsofIvy
#4
Sep21-03, 10:18 AM
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3d projectile motion

Then apparently you don't understand your own problem. You said originally, "You are driving in a car moving at 52 km/h [E]. "

In your second post you said "the car is moving on the road. Therefore, 22 km/h is the velocity of the car relative to the ground."

Which is it? 52 km/hr or 22 km/hr?


You originally said "YOu thrown an empty pop can out the window at 22km/h [N 30 degrees W] at an angle of 15 degrees above the horizontal."

The question of whether that is relative to the road or to the car is very important and if you don't see that you need to go back and read your textbook again.

Oh, and here's another question: what's the fine for throwing a pop can out the car window?
thegame
#5
Sep21-03, 11:04 AM
P: 32
Originally posted by HallsofIvy
Then apparently you don't understand your own problem. You said originally, "You are driving in a car moving at 52 km/h [E]. "

In your second post you said "the car is moving on the road. Therefore, 22 km/h is the velocity of the car relative to the ground."

Which is it? 52 km/hr or 22 km/hr?

my bad, the car 52 km/h [e] relative to the ground
The pop can is thrown out at 22 km/h [N 30 degrees W] relative to the car...

BTW.. it said the pop can was being amied at a garbage can but I saved myself some typing
HallsofIvy
#6
Sep21-03, 11:53 AM
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I sure hope you hit that garbage can!

The car is moving at 52 km/hr due E.
The pop can is thrown at 22 km/hr at 30 degrees west of north.

We might set up a coordinate system with "x" to the east and "y" north. Then the car's velocity can be written as the vector <52, 0>.

The velocity of the pop can, relative to the car can be written as the vector <-22 cos(30),22 sin(30)>= <-11, 11[sqrt](3)>.

The velocity of the pop can, relative to the ground can be written as the sum of those: <41, 11[sqrt](3)>.

NOW, calculate the speed of the pop can- the "length" of that vector- and use that as initial speed in your standard solution for parabolic motion: y= -(g/2)t2+ v0 cos(15) t and
x= v0 sin(15)t. Determine what t is when y= 0 again, put that into the equation for x to find the distance away from the intial point (on the road) the can lands. Since the can continues on the line of the velocity vector, multiply that vector, <-11, 11[sqrt](3)>, by the appropriate factor to give the correct length. That resulting vector will give the position at which the can hits.


Of course, since those smart-*** teenagers in the car forgot to allow for wind-resistance (VERY important for an empty pop can), the can will miss the garbage can and hit the highway patrolman sitting on his motorcycle behind the garbage can. The highway patrolman will give them a stiff fine for littering (not to mention assaulting a policeman!) and their parents will ground them FOR LIFE- thus making our highways safer for us all! Lovely problem!
thegame
#7
Sep21-03, 01:52 PM
P: 32
In class we only learned projectile motion in 2 dimensional so what you did above is very hard to understand. Can someone explain it using 2d projectile motion where you put all the x direction on one side and all the y direction information on the other side.
thegame
#8
Sep21-03, 01:56 PM
P: 32
Originally posted by HallsofIvy
The velocity of the pop can, relative to the car can be written as the vector <-22 cos(30),22 sin(30)>= <-11, 11[sqrt](3)>.
dont understand this part
HallsofIvy
#9
Sep21-03, 09:16 PM
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In class we only learned projectile motion in 2 dimensional so what you did above is very hard to understand. Can someone explain it using 2d projectile motion where you put all the x direction on one side and all the y direction information on the other side.
This IS in 2 dimensions. I DID put "all the x direction on one side and all the y direction information on the other side" (of the vector. If that's not what you mean by "one side" and "the other side", then I don't know what you want.


Suppose a projectile is launched at angle theta to the horizontal with intial speed vo. What are the equations for the horizontal and vertical components of displacement?


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