Physics problem

[Blaine0002]

I have no clue how to figure this out, and its hard to know what to google for.
If anyone could explain how to get an answer it would be greatly appreciated

I shoot something at an angle of 60 degrees, and it travels 15 miles, what is the initial velocity of the projectile?

Thanks!

ya know, the Community Lounge... better place to post this.

and the initial velocity is zero... initial means at the start... the object is not moving. (i am always the one to break the party)

[Paul M]

Originally Posted by nexialys

ya know, the Community Lounge... better place to post this.

Agreed and moved.

[Opserty]

Originally Posted by nexialys

and the initial velocity is zero... initial means at the start... the object is not moving. (i am always the one to break the party)

In Kinematics initial doesn't always refer to the very start, you can take the initial velocity of an object when its already moving and use that as your starting point.

Anyhoo...with regards to your problem I think you have to make a point of the horizontal component being independent of the vertical component. The object will travel in a parabola.

Vertically speaking:
Displacement = 0 (Your going up then coming back down)
Initial velocity = Vertical component of Initial velocity. ( u sin60)
Acceleration = Acceleration due to gravity [-9.8 (+ve or -ve depending on your frame of reference)]
Time = Unknown
[S]Final Velocity = Not interested[/S]

Horizontally:
Displacement = 15 miles (Quite why your still working with imperial units I don't know :erm
Initial velocity = Horizontal component of Initial velocity. ( u cos60 = 0.5u)
Acceleration = 0 (Assuming there is no air resistance)
Time = Unknown
[S]Final Velocity = Not interested[/S]

Then for both I used this equation:


Solve it for Horizontal first, that will give you a value of t in terms of u. Substitute this into the same equation for the vertical motion and it will give you the Initial Horizontal Velocity.

Don't forget to calculate the Initial Velocity from its Horizontal component, (basically its u/cos60)

That is how I would tackle it anyway, bearing in mind I haven't done any kind of projectile motion, so there may be an easier way to do it.

Good luck

[UncoderMom]

Originally Posted by Opserty

In Kinematics initial doesn't always refer to the very start, you can take the initial velocity of an object when its already moving and use that as your starting point.

Anyhoo...with regards to your problem I think you have to make a point of the horizontal component being independent of the vertical component. The object will travel in a parabola.

Vertically speaking:
Displacement = 0 (Your going up then coming back down)
Initial velocity = Vertical component of Initial velocity. ( u sin60)
Acceleration = Acceleration due to gravity [-9.8 (+ve or -ve depending on your frame of reference)]
Time = Unknown
[S]Final Velocity = Not interested[/S]

Horizontally:
Displacement = 15 miles (Quite why your still working with imperial units I don't know :erm
Initial velocity = Horizontal component of Initial velocity. ( u cos60 = 0.5u)
Acceleration = 0 (Assuming there is no air resistance)
Time = Unknown
[S]Final Velocity = Not interested[/S]

Then for both I used this equation:


Solve it for Horizontal first, that will give you a value of t in terms of u. Substitute this into the same equation for the vertical motion and it will give you the Initial Horizontal Velocity.

Don't forget to calculate the Initial Velocity from its Horizontal component, (basically its u/cos60)

That is how I would tackle it anyway, bearing in mind I haven't done any kind of projectile motion, so there may be an easier way to do it.

Good luck

Yeah that!

[Boofo]

No, that looks right to me.

[Blaine0002]

Originally Posted by Opserty

In Kinematics initial doesn't always refer to the very start, you can take the initial velocity of an object when its already moving and use that as your starting point.

Anyhoo...with regards to your problem I think you have to make a point of the horizontal component being independent of the vertical component. The object will travel in a parabola.

Vertically speaking:
Displacement = 0 (Your going up then coming back down)
Initial velocity = Vertical component of Initial velocity. ( u sin60)
Acceleration = Acceleration due to gravity [-9.8 (+ve or -ve depending on your frame of reference)]
Time = Unknown
[S]Final Velocity = Not interested[/S]

Horizontally:
Displacement = 15 miles (Quite why your still working with imperial units I don't know :erm
Initial velocity = Horizontal component of Initial velocity. ( u cos60 = 0.5u)
Acceleration = 0 (Assuming there is no air resistance)
Time = Unknown
[S]Final Velocity = Not interested[/S]

Then for both I used this equation:


Solve it for Horizontal first, that will give you a value of t in terms of u. Substitute this into the same equation for the vertical motion and it will give you the Initial Horizontal Velocity.

Don't forget to calculate the Initial Velocity from its Horizontal component, (basically its u/cos60)

That is how I would tackle it anyway, bearing in mind I haven't done any kind of projectile motion, so there may be an easier way to do it.

Good luck

Thanks alot for your help, ill need to study this!

[iogames]

Originally Posted by Opserty

In Kinematics initial doesn't always refer to the very start, you can take the initial velocity of an object when its already moving and use that as your starting point.

Anyhoo...with regards to your problem I think you have to make a point of the horizontal component being independent of the vertical component. The object will travel in a parabola.

Vertically speaking:
Displacement = 0 (Your going up then coming back down)
Initial velocity = Vertical component of Initial velocity. ( u sin60)
Acceleration = Acceleration due to gravity [-9.8 (+ve or -ve depending on your frame of reference)]
Time = Unknown
[S]Final Velocity = Not interested[/S]

Horizontally:
Displacement = 15 miles (Quite why your still working with imperial units I don't know :erm
Initial velocity = Horizontal component of Initial velocity. ( u cos60 = 0.5u)
Acceleration = 0 (Assuming there is no air resistance)
Time = Unknown
[S]Final Velocity = Not interested[/S]

Then for both I used this equation:


Solve it for Horizontal first, that will give you a value of t in terms of u. Substitute this into the same equation for the vertical motion and it will give you the Initial Horizontal Velocity.

Don't forget to calculate the Initial Velocity from its Horizontal component, (basically its u/cos60)

That is how I would tackle it anyway, bearing in mind I haven't done any kind of projectile motion, so there may be an easier way to do it.

Good luck

I concur!

[Opserty]

I'm glad I have the support of some of the forum! lol

P.S. Blaine002 if it doesn't workout try here: http://www.physicsforums.com/ I found 'em through vb.org

Wow they make a forum for any school subject.

Anyway, like Opserty said, the initial velocity would be from where ever you started calculating the distance, therefore, you could think of it as 60 degrees being the angle of elevation and the 15 miles being the ground distance. That is just taking it from a mathematical point of view. I bet there is someone else who has an actual answer.

-- Ling Ling