Titled reference frame, N2L with position and velocity

[learningphysics]

will i be setting vy to zero and then solving a free fall?

I don't understand... the question asks you to find the range in the x direction I believe...

 

[Oblio]

using kinematics?

 

[learningphysics]

using kinematics?

yes... what is the displacement along the x direction and y direction... in terms of time...

 

[Oblio]

d = v0*t + (1/2)at^2

 

[learningphysics]

d = v0*t + (1/2)at^2

Yes. Now use the values we've calculated for v0 and a for this problem... for the x- direction... and the y-direction...

 

[Oblio]

dx = vocos$$\vartheta$$t + (1/2)(-g)t^2sin$$\phi$$

dy = vosin$$\vartheta$$t + (1/2)(-g)t^2cos$$\phi$$

 

[Oblio]

i have a feeling this is wrong for dy...

 

[learningphysics]

i have a feeling this is wrong for dy...

Looks right to me. So now you want to find the range...

 

[Oblio]

range?

 

[learningphysics]

range?

the question asks for the range...

 

[Oblio]

which is really just the displacement up the plane?

 

[learningphysics]

which is really just the displacement up the plane?

yes, maximum displacement... the object is thrown at an angle, then hits the incline eventually... the displacement up the plane when it hits is the range...

 

[Oblio]

is this the same kinematics equation?

 

[learningphysics]

dx = vocos$$\vartheta$$t + (1/2)(-g)t^2sin$$\phi$$

dy = vosin$$\vartheta$$t + (1/2)(-g)t^2cos$$\phi$$

use these equations to find dx when dy = 0... that gives the range.

 

[Oblio]

solve for... vo? and insert into dx?

 

[learningphysics]

solve for... vo? and insert into dx?

no, you want to eliminate t... so solve for t and substitute into dx.

 

[Oblio]

without simplifying yet..

dx = vocos (theta) (vosin (theta) / (-1/2)gsin(phi)) + (1/2)(-g)cos(phi)((vosin(theta)/(-1/2)gsin)^2)

ya?

 

[learningphysics]

without simplifying yet..

dx = vocos (theta) (vosin (theta) / (-1/2)gsin(phi)) + (1/2)(-g)cos(phi)((vosin(theta)/(-1/2)gsin)^2)

ya?

hmm... close but some mistakes... can you post what you got for t?

 

[Oblio]

t = vosin(theta) / (-1/2)gsin(phi)

 

[learningphysics]

t = vosin(theta) / (-1/2)gsin(phi)

setting dy = 0, and solving for t will give you something different... it should be cos(phi) and the - shouldn't be there.

 

[Oblio]

oops! little pen and ink mistake over here..

t = vosin(theta) / (1/2)gcos(phi)
yep!

 

[learningphysics]

oops! little pen and ink mistake over here..

t = vosin(theta) / (1/2)gcos(phi)
yep!

cool. plugging that into dx should work.

 

[Oblio]

dx = vocos(theta)vosin(theta)/(1/2)gcos(phi) - (1/2)gsin(phi)vosin(theta)/(1/2)gcos(phi)

 

[learningphysics]

dx = vocos(theta)vosin(theta)/(1/2)gcos(phi) - (1/2)gsin(phi)vosin(theta)/(1/2)gcos(phi)

you missed t^2... didn't square t.

 

[Oblio]

oops.
man I am bad..

 

[Oblio]

dx = vocos(theta)vosin(theta)/(1/2)gcos(phi) - (1/2)gsin(phi)(vosin(theta)/(1/2)gcos(phi))^2

on the right side i can cancel out the (1/2) on the bottom and top, as well as the g, giving,

dx = vocos(theta)vosin(theta)/(1/2)gcos(phi) - sin(phi)(vosin(theta)/cos(phi))^2

 

[learningphysics]

dx = vocos(theta)vosin(theta)/(1/2)gcos(phi) - (1/2)gsin(phi)(vosin(theta)/(1/2)gcos(phi))^2

on the right side i can cancel out the (1/2) on the bottom and top, as well as the g, giving,

dx = vocos(theta)vosin(theta)/(1/2)gcos(phi) - sin(phi)(vosin(theta)/cos(phi))^2

careful the 1/2 and g are squared...

 

[Oblio]

oops again.

so I am left with
dx = vocos(theta)vosin(theta)/(1/2)gcos(phi) - sin(phi)(vo^2 sin(theta) ^2 / (1/2) g (cos(phi)^2)

 

[learningphysics]

oops again.

so I am left with
dx = vocos(theta)vosin(theta)/(1/2)gcos(phi) - sin(phi)(vo^2 sin(theta) ^2 / (1/2) g (cos(phi)^2)

Looks good... try to simplify and use a trig identity to get it to look like the formula they gave for the range...

 

[Oblio]

will trig identities apply since theta and phi are present?