Problem regarding projectile motion on a inclined plane

[CherryWine]

Homework Statement

A projectile is thrown on an inclination plane with incline angle phi with an inital velocity V0 on an angle theta. Find the range of the projectile; treat the hypotenuse of the inclined plane as the x-axis. For what value of initial velocity angle is the range maximum, and what is that range.

Homework Equations

All 2D kinematics equations.

The Attempt at a Solution

Firstly I used the equation y(t) and expressed flight time as 2Vy/-gy (g is negative itself, the minus in front of it makes it positive), from there I wrote t=(2V0*sintheta)/(-g*cosphi). Then I plugged that into the equation for range which is D=Vx*t+1/2gx*t^2 where gx is the x-axis component of g. By simplifying the expression I obtained D=(2V0^2*sintheta*(-costheta*cosphi+sintheta*sinphi))/(g*(cosphi)^2). Now everything is correct except the bolded part which should be costheta*sin(theta-phi). Help please.

 

[TSny]

Note that the angle θ in the figure is measured from the horizontal, not from the inclined surface. It appears that you used V_y = V₀ sinθ. Is this correct?

(If you click on the Σ symbol on the formatting tool bar, you can access more math symbols, like θ.)

 

[CherryWine]

Note that the angle θ in the figure is measured from the horizontal, not from the inclined surface. It appears that you used V_y = V₀ sinθ. Is this correct?

(If you click on the Σ symbol on the formatting tool bar, you can access more math symbols, like θ.)

Yes! Sorry for the careless mistake, I found another attempt using the correct equation for t, but still everything is correct except the trigonometric part which then equals to sin(θ-Φ)*(sin(θ-Φ)*sinθ-cos(θ-Φ)*cosΦ), and I used Wolfram Alpha to check if this was equal to the correct cosθ*sin(θ-Φ) and it was not.

 

[TSny]

Yes! Sorry for the careless mistake, I found another attempt using the correct equation for t, but still everything is correct except the trigonometric part which then equals to sin(θ-Φ)*(sin(θ-Φ)*sinθ-cos(θ-Φ)*cosΦ), and I used Wolfram Alpha to check if this was equal to the correct cosθ*sin(θ-Φ) and it was not.

You have sin(θ-Φ)*(sin(θ-Φ)*sinθ-cos(θ-Φ)*cosΦ). I suspect that the mistake is the angle θ in red. Check that.

 

[CherryWine]

You have sin(θ-Φ)*(sin(θ-Φ)*sinθ-cos(θ-Φ)*cosΦ). I suspect that the mistake is the angle θ in red. Check that.

Yes! Thank you! It should be sinΦ. Thanks!

 

[CherryWine]

You have sin(θ-Φ)*(sin(θ-Φ)*sinθ-cos(θ-Φ)*cosΦ). I suspect that the mistake is the angle θ in red. Check that.

Could you please prove algebraically that sin(θ-Φ)*(sin(θ-Φ)*sinΦ-cos(θ-Φ)*cosΦ)=cosθ*sin(θ-Φ), and also if you could give me some hints for finding the angle for which that expression is going to be maximum. Thank you.

 

[TSny]

Could you please prove algebraically that sin(θ-Φ)*(sin(θ-Φ)*sinΦ-cos(θ-Φ)*cosΦ)=cosθ*sin(θ-Φ)

Let α = θ-Φ and β = Φ. Write sin(θ-Φ)*sinΦ-cos(θ-Φ)*cosΦ in terms of α and β. Then see if it reminds you of a well-known trig identity.

and also if you could give me some hints for finding the angle for which that expression is going to be maximum. Thank you.

Write D in the form D = C⋅f(θ), where C is independent of θ. D is maximized when f(θ) is maximized

 

[CherryWine]

Let α = θ-Φ and β = Φ. Write sin(θ-Φ)*sinΦ-cos(θ-Φ)*cosΦ in terms of α and β. Then see if it reminds you of a well-known trig identity.Write D in the form D = C⋅f(θ), where C is independent of θ. D is maximized when f(θ) is maximized

Thanks for the hint, I proved it. If you could just tell me what would you do to find the angle for maximum f(θ) because my brain is really blocking currently. And if you can, do not use derivatives.

 

[TSny]

If you could just tell me what would you do to find the angle for maximum f(θ) because my brain is really blocking currently. And if you can, do not use derivatives.

Try using another trig identity: http://www.sosmath.com/trig/prodform/prodform.html