Need help by 10:00 tonight (when assignment is due)

[Bensky]

Homework Statement

http://xs221.xs.to/xs221/07452/graphhelp.PNG

Find the initial x velocity and initial y velocity.

Homework Equations

vx0=vx

The Attempt at a Solution

I tried putting 0 as both, but later realized that it has to have some initial x/y velocity or the graph would not curve upwards. I think that the initial y acceleration has to be negative because of the parabolic shape, but I don't know what the velocity is. :$ As for the initial x velocity, all I can figure out is that its constant.

 

[Kurdt]

I'm not sure what it is you're trying to do. The statement of the problem isn't very clear. Could you restate it exactly as its given, or if that is exactly as its given could you elaborate more about what you're doing.

 

[Bensky]

I'm not sure what it is you're trying to do. The statement of the problem isn't very clear. Could you restate it exactly as its given, or if that is exactly as its given could you elaborate more about what you're doing.

Sorry, here is the whole problem (the graph was at the top and was an animation):

Problem: An object is launched near the Earth as shown in the animation.

So far, I have figured out the following values (they are correct, so the computer says). The ones with ?s are the ones that I don't know.
What is the maximum height of the object? 26 m
What is the initial x velocity? ? m/s

What is the initial y velocity? ? m/s

What is the speed of the object at the maximum height? 10.6 m/s

What is the acceleration of the object at the maximum height? -9.8 m/s2

 

[Kurdt]

Unfortunately its not animated on here so we don't know what the graph looks like at all. Is it a distance-time graph or are both axes distance, and what does it look like?

 

[Bensky]

Unfortunately its not animated on here so we don't know what the graph looks like at all. Is it a distance-time graph or are both axes distance, and what does it look like?

Both axes are distance. The shape of the graph is like y=-x^2 (parabola, curves at the top then comes downward)
As it animates, the time at the top increases :$

is that what you needed?

 

[Bensky]

Oh, and here is a picture of the graph at the end:
http://xs221.xs.to/xs221/07452/endofgraph.PNG

 

[Kurdt]

The curve is symmetrical which means just before the projectile hits the ground its speed in the x and y directions are linked to the initial x and y speeds. Can you make the connection?

 

[Bensky]

The curve is symmetrical which means just before the projectile hits the ground its speed in the x and y directions are linked to the initial x and y speeds. Can you make the connection?

I can see the ending velocity will be the same as the beginning velocity. Do I need to use a kinematic equation or do I need to find the slope to get the initial x and y velocities? :\

I know this is easy but I can't seem to get it.

 

[Kurdt]

One can use the kinematic equations. Obviously x is constant as you stated and is easy to work out from the info on the graph. The y-component can be found by using the kinematic equations as I said. You know the maximum height the starting speed which will be zero m/s and the acceleration which is just -g and finally the time which will be half the total time.

Be careful about signs though. The initial velocity in the y-direction is in a different direction to the final velocity.

 

[Bensky]

One can use the kinematic equations. Obviously x is constant as you stated and is easy to work out from the info on the graph. The y-component can be found by using the kinematic equations as I said. You know the maximum height the starting speed which will be zero m/s and the acceleration which is just -g and finally the time which will be half the total time.

Be careful about signs though. The initial velocity in the y-direction is in a different direction to the final velocity.

So the initial x velocity would be 10.6 since that's what it is at the top?

 

[Kurdt]

So the initial x velocity would be 10.6 since that's what it is at the top?

Yes, you can verify that with the info given in the graph.

 

[Bensky]

Alright, thank you, I understand now. :)