(projectile motion) calculation problem

[asdf1]

Question: A ball with a velocity of Vo is thrown from the bottom of an inclined plane (point O) with an angle of \phi. The inclined plane has an angle oftheta . The ball lands on top of the inclined plane on a point A. Find the length of OA and its maximum value.

My calculations:
1. x = V_{0}\cos (\phi) t
2. y=Vo*sin\phit -0.5*gt^2
3. y=xtan\theta

From the above equations, I got t=[2Vo*sin(\phi-\theta)/[gcos\theta]
So OA= xsec\theta
=Vo*cos\phi*sec\theta/(gcos\theta)

But the answer is
OA= Vo^2*[sin(2\phi-\theta)-sin\theta]/[gcos^2(\theta)]

And I don't know how to tell from my calculations what the max value is, and why it's different from the answer...
Can somebody help?

http://img97.imageshack.us/my.php?image=inclinedplanesu8.png"

 

[Hootenanny]

Change your tags to [ tex ] ... [ /tex ] (without the spaces obviously)

Click on this equation below to see how it is coded;

x = V_{0}\cos (\phi) t

 

[asdf1]

thanks for teaching me latex! ^_^

 

[Hootenanny]

1. x = V_{0}\cos (\phi) t
2. y=Vo*sin\phit -0.5*gt^2
3. y=xtan\theta

From the above equations, I got t=[2Vo*sin(\phi-\theta)/[gcos\theta]

How did you arrive at this? Would you mind posting your steps?

 

[asdf1]

from 1. and 3. ->
y=V_{0}\cos (\phi) t*tan\theta
then take the above result and plug it into 2. getting:
y=Vo*sin\phit -0.5*gt^2=V_{0}\cos (\phi) t*tan\theta

Rearranging the equation and using the formula
sin2\phi=2sin\phicos\phi to simplfy the equation makes the result for
t=[2Vo*sin(\phi-\theta)/[gcos\theta]

 

[pizzasky]

From the above equations, I got t=[2Vo*sin(\phi-\theta)/[gcos\theta]
So OA= xsec\theta

Up to this point, everything is fine.

OA=Vo*cos\phi*sec\theta/(gcos\theta)

You seem to have made an error here. What is the expression for x?

 

[asdf1]

Sorry! I mistyped!
It should be
OA= Vo*cos\phi*2Vo*sin(\phi-\theta)*sec\phi/[gcos\phi]

 

[pizzasky]

OA= Vo*cos\phi*2Vo*sin(\phi-\theta)*sec\phi/[gcos\phi]

Do you mean OA= Vo*cos\phi*2Vo*sin(\phi-\theta)*sec\theta/[gcos\theta]?

If yes, then you are on the right track. Compare your answer with the one given in the book. What is the last step you need to take?

 

[asdf1]

Yes...
I think that I have to use some kind of trignometric formula to get the answer in the book?

 

[leo_thunderbird]

the ans should come out to be
R=(2u<sq (only u)>sin<sq>(a-b)cos<sq>a)/(gcosb)<sq>

the max value will be when a= 45-b/2

where a = angle of projection + angle of inclination(b)
angle of inclination=b

if u ask i will mail u the proof

 

[ambuj123]

A Better solution

Hello
well i have written a different solution
i am taking axis along the incline and perpendicular to the incline then taking components of g(acceleration due to gravity) along and perpendicular to the incline and then applying kinematics equations
anywayshttp://helpjee.blogspot.com/2006/08/inclined-plane-problem_01.html"
bye

 

[asdf1]

Wow! thank you for the different approach!