How can I solve the proof for a projectile fired up an incline?

[Epiphone]

Homework Statement

A projectile is fired up an incline (incline angle φ) with an initial speed vi at an angle θi with respect to the horizontal (θi > φ). (a.) Show that the projectile travels a distance d up the incline, where

d = 2*vi^2*cosθi*sin(θi-φ) / g*cos(φ)^2

Homework Equations

$$v = v_0 + a t$$

The Attempt at a Solution

I have tried 2 different approaches to this proof. The first strategy was to tilt the figure so that the distance d was parallel to the x axis. However, through some research I found that to rotate the problem, you will also need to find the components of gravity and things got hairy from there.

The second approach I took was to write the normal equations for x and y

( x= v_i cos (\theta_i) t, y= v_i sin ( \theta_i) t - 1/2 g t^2 ).

Then I plugged x= d cos (\phi) and y = d sin ( \phi) into the 2 above equations

I ended up with d(cos(phi)) = v_i(cos(θi))t
and d(sin(phi)) = v_i(sin(θi)t - .5gt^2

I was then told to isolate the t value from the first equation, and plug it into the second equation.

However, this resulted in a very complex equation and I was unable to solve it for d, which should have given me the initial equation I was trying to prove.

Am I making any errors, or should I try a different method?

Thanks

 

[learningphysics]

your second appraoch seems fine. when you isolate t from this equation:

d(cos(phi)) = v_i(cos(θi))t

and substitute it into the second... you should be able to solve for d... a d cancels on both sides of the equation... so it should be simple to solve for d...

 

[Epiphone]

i see what you mean, but i am having trouble manipulating the variables to isolate d. so far i have:

d(sin(phi)) =(visin(θi)vicos(θi))/dcos(phi)-(gvi^2cos^2(θi))/2d^2cos^2(θ))

 

[learningphysics]

i see what you mean, but i am having trouble manipulating the variables to isolate d. so far i have:

d(sin(phi)) =(visin(θi)vicos(θi))/dcos(phi)-(gvi^2cos^2(θi))/2d^2cos^2(θ))

how are you getting d in the denominator? t = dcos(phi)/vicos(θi)

 

[Epiphone]

oh wow, you're absolutely right
let me re-try this

 

[Epiphone]

my second attempt definitely did not work. somehow i lost all of my "v"s

I started with:
dsin(phi) = vsin(θ)dcos(phi)/vcos(θ) - .5g((dcos(phi)/vcos(θ))^2

and ended up with:
d = (sin(phi) - tan(θ)cos(phi)) / (-gcos^2(phi))

should i factor out the dcos(phi)/vcos(θ) initially?

 

[learningphysics]

my second attempt definitely did not work. somehow i lost all of my "v"s

I started with:
dsin(phi) = vsin(θ)dcos(phi)/vcos(θ) - .5g((dcos(phi)/vcos(θ))^2

the v's cancel in your first term on the right hand side... there's a v^2 in the denominator in the second term on the right hand side.

 

[Epiphone]

ok, I'm pretty sure I've done all the math correctly, but I've gotten stuck again

i've gotten d by itself:

d = [2v^2cos^2(θ)sin(phi)/-gcos^2(phi)] - [2v^2cos^2(θ)sin(θ)cos(phi)/-gcos^2(phi)cos(θ)]

i am unsure how to get this into the original form, or if i made a mistake along the way

 

[learningphysics]

ok, I'm pretty sure I've done all the math correctly, but I've gotten stuck again

i've gotten d by itself:

d = [2v^2cos^2(θ)sin(phi)/-gcos^2(phi)] - [2v^2cos^2(θ)sin(θ)cos(phi)/-gcos^2(phi)cos(θ)]

i am unsure how to get this into the original form, or if i made a mistake along the way

use sin(A-B) = sinAcosB - cosAsinB

 

[Epiphone]

it took a lot of reviewing to find all the little errors i made, but i finally got it to work out
thank you very much for all of your help.

im glad you gave me that trig identity, I've honestly never learned any of them. perhaps in precalc last year, but I've never used any.

thanks again

 

[laforge]

Hi, I hope it's ok that I am brining this thread back but I am having some problems with this proof.

I started with the same equations he has

$$dsin\phi = (vsin \thetadcos \phi )/(vcos \theta ) - (.5gd^2cos^2 \phi )/(v^2cos^2 \theta )$$

and I reduce it to

$$d=(2sin \theta cos \phi v^2cos \theta -2sin \phi v^2cos^2 \theta )/(gcos^2 \phi)$$

Which is not what the original poster got. As you can see my terms on the right are reversed and I have a $$cos^2 \theta$$ in one and he doesn't.

I have gone through the math quite a few times and I can't see an error.

Could anyone think of a possible error that I could be having? I will post my whole work if needed but it is quite long.

 

[learningphysics]

Hi, I hope it's ok that I am brining this thread back but I am having some problems with this proof.

I started with the same equations he has



and I reduce it to



Which is not what the original poster got. As you can see my terms on the right are reversed and I have a $$cos^2 \theta$$ in one and he doesn't.

I have gone through the math quite a few times and I can't see an error.

Could anyone think of a possible error that I could be having? I will post my whole work if needed but it is quite long.

Looks the same to me, comparing with post #8.