# Problem regarding projectile motion on a inclined plane

• CherryWine
In summary, the conversation discusses finding the range of a projectile thrown on an inclined plane, as well as determining the initial velocity angle for maximum range. The solution involves using various kinematics equations and trigonometric identities.
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.

Note that the angle θ in the figure is measured from the horizontal, not from the inclined surface. It appears that you used Vy = 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.)

TSny said:
Note that the angle θ in the figure is measured from the horizontal, not from the inclined surface. It appears that you used Vy = 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.

CherryWine said:
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
TSny said:
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!

TSny said:
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.

CherryWine said:
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
TSny said:
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.

CherryWine said:
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

## 1. What is projectile motion on an inclined plane?

Projectile motion on an inclined plane refers to the motion of an object that is launched at an angle on a surface that is not horizontal. This type of motion is characterized by both horizontal and vertical components of velocity.

## 2. How does the angle of inclination affect the projectile motion?

The angle of inclination, or the angle at which the inclined plane is tilted, can affect the range, height, and flight time of the projectile. A steeper angle will result in a shorter range and higher flight time, while a shallower angle will result in a longer range and shorter flight time.

## 3. What is the formula for calculating the range of a projectile on an inclined plane?

The formula for calculating the range of a projectile on an inclined plane is R = (v2sin2θ)/g, where R is the range, v is the initial velocity, θ is the angle of inclination, and g is the acceleration due to gravity.

## 4. How does the acceleration due to gravity affect the projectile motion on an inclined plane?

The acceleration due to gravity affects the vertical component of the projectile's motion. As the angle of inclination increases, the vertical component of the projectile's velocity decreases, which also affects the range and flight time of the projectile.

## 5. Can the initial velocity of the projectile affect its motion on an inclined plane?

Yes, the initial velocity of the projectile can affect its motion on an inclined plane. A higher initial velocity will result in a longer range and shorter flight time, while a lower initial velocity will result in a shorter range and longer flight time.

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