# (projectile motion) calculation problem

1. Jul 29, 2006

### 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 of$$theta$$ . 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$$\phi$$t -0.5*g$$t^2$$
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$$\theta$$)

OA= $$Vo^2$$*[sin$$(2\phi-\theta)$$-sin$$\theta$$]/[g$$cos^2(\theta)$$]

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"

Last edited by a moderator: Apr 22, 2017
2. Jul 29, 2006

### Hootenanny

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

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

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

Last edited: Jul 29, 2006
3. Jul 29, 2006

### asdf1

thanks for teaching me latex! ^_^

4. Jul 29, 2006

### Hootenanny

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

5. Jul 30, 2006

### 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$$\phi$$t -0.5*g$$t^2$$=$$V_{0}\cos (\phi) t$$*tan$$\theta$$

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

6. Jul 31, 2006

Up to this point, everything is fine.
You seem to have made an error here. What is the expression for x?

Last edited: Jul 31, 2006
7. Jul 31, 2006

### asdf1

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

8. Jul 31, 2006

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?

Last edited: Jul 31, 2006
9. Aug 1, 2006

### asdf1

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

10. Aug 1, 2006

### 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

11. Aug 1, 2006

### 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" [Broken]
bye

Last edited by a moderator: May 2, 2017
12. Aug 3, 2006

### asdf1

Wow! thank you for the different approach!