Projectile thrown up an incline, max distance?

Since \varphi is the same as in the original equation, I'm assuming the R_max in the original problem is incorrect. If so, you don't have to worry about maximizing it.In summary, the distance a ball travels up an inclined plane is R = (2V^2sinθcos(θ+Φ))/(gcos^2Φ), and the maximum possible range up the inclined plane is R_max = V^2/(g(1+sinΦ)).
  • #1
sabinscabin
11
0

Homework Statement



A ball is thrown with initial speed [tex] V [/tex] up an inclined plane. The plane is inclined at an angle [tex]\phi[/tex] above the horizontal, and the ball's velocity is at an angle [tex]\theta[/tex] above the plane.

Show that the ball lands a distance [tex]R = \frac{ 2V^2 \sin{\theta} \cos{ ( \theta + \phi} ) } { g \cos^2{\phi} }[/tex] from its launch point. Show that for a given [tex]V[/tex] and
[tex]\phi[/tex], the maximum possible range up the inclined plane is
[tex] R_{max} = \frac{ V^2 }{ g (1 + \sin{\phi} )} [/tex]

Homework Equations



F = ma

The Attempt at a Solution



I calculated the distance traveled up the incline fine. However, I'm having trouble proving the second part. I'm guessing I'm supposed to maximize R with respect to theta, so from the equation above we have:

[tex]\frac{d} {d \theta} \sin{\theta} \cos{(\theta + \phi)} = 0[/tex]

[tex]\cos{( 2 \theta + \phi )} = 0[/tex]

[tex]\theta = \frac{n \pi}{4} - \frac{\phi}{2} [/tex], with n = odd integer. Now plug this back into the original equation for R and I get

[tex]R_{max} = \frac{ 4V^2 \tan{\phi} }{g} \cos{ 2\phi} [/tex]

I think I'm approaching this problem wrong. Can anyone give me a simpler way?
 
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  • #3
Bright Wang said:

if you look at the last post in the thread you gave me, I already got that result. It's just that the R_max I get does not match the R max that's given in the initial problem.
 
  • #4
sabinscabin said:
if you look at the last post in the thread you gave me, I already got that result. It's just that the R_max I get does not match the R max that's given in the initial problem.

Ops my bad.
 
  • #5
sabinscabin said:
[tex]\theta = \frac{n \pi}{4} - \frac{\phi}{2} [/tex], with n = odd integer. Now plug this back into the original equation for R and I get

[tex]R_{max} = \frac{ 4V^2 \tan{\phi} }{g} \cos{ 2\phi} [/tex]

I think I'm approaching this problem wrong. Can anyone give me a simpler way?

First, n must be 1; if it's 3 or higher, than n*pi/4 - Φ/2 would be greater than 90 degrees, and that's impossible given the geometry of the situation. So we know that the optimum angle is pi/4-Φ/2. Substituting it into the equation for R gives:

R=sin(pi/4-Φ/2)cos(pi/4+Φ/2)/cos^2 (Φ)

I've ignored the constants. Since cos(a)=sin(pi/2-a):

R=sin(pi/4-Φ/2)sin(pi/4-Φ/2)
cos(2a)=1-2sin^2 (a), so

R=(1-cos(pi/2-Φ))/2
=(1-sin(Φ))/2

You seem to have decent math skills, so getting from here to the answer should be easy.
 
  • #6
you have got [tex]\theta[/tex] = [tex]\pi/4[/tex] - [tex]\varphi/2[/tex]
so 2sin[tex]\theta[/tex] cos([tex]\theta[/tex] + [tex]\varphi[/tex]) = 2sin([tex]\pi/4[/tex] - [tex]\varphi/2[/tex]) sin([tex]\pi/4[/tex] - [tex]\varphi/2[/tex]) =
2sin2([tex]\pi/4[/tex] - [tex]\varphi/2[/tex]) = [cos([tex]\varphi/2[/tex]) - sin([tex]\varphi/2[/tex])]2

[cos([tex]\varphi/2[/tex]) - sin([tex]\varphi/2[/tex])]2/cos2[tex]\varphi[/tex]
= {[cos([tex]\varphi/2[/tex]) - sin([tex]\varphi/2[/tex])]/cos[tex]\varphi[/tex]}2
= {[cos([tex]\varphi/2[/tex]) - sin([tex]\varphi/2[/tex])]/[cos2([tex]\varphi/2[/tex]) - sin2([tex]\varphi/2[/tex])]}2
= {1/[cos([tex]\varphi/2[/tex]) + sin([tex]\varphi/2[/tex])]}2
= 1/(1 + sin[tex]\varphi[/tex])
 

1. What is a projectile thrown up an incline?

A projectile thrown up an incline is an object that is thrown or launched with an initial velocity at an angle from a surface that has an upward slope.

2. How does the incline affect the distance of a projectile thrown up an incline?

The incline will affect the distance of the projectile by changing the angle at which it is thrown and the gravitational pull acting on it. This will result in a curved path and a shorter horizontal distance compared to throwing the projectile on a flat surface.

3. What is the maximum distance that a projectile thrown up an incline can travel?

The maximum distance that a projectile thrown up an incline can travel is called the range. It is the furthest horizontal distance that the projectile can travel before hitting the ground.

4. How can the maximum distance of a projectile thrown up an incline be calculated?

The maximum distance of a projectile thrown up an incline can be calculated using the range formula: R = (v^2 * sin2θ) / g, where R is the range, v is the initial velocity, θ is the angle at which the projectile is thrown, and g is the acceleration due to gravity.

5. What factors can affect the maximum distance of a projectile thrown up an incline?

The factors that can affect the maximum distance of a projectile thrown up an incline include the initial velocity, angle of projection, air resistance, and the incline height and slope. Other external factors like wind and friction can also affect the distance traveled.

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