Projectile Motion: Launching Up a Hill

Maximizing this equation with respect to α, we get α = β/2 + π/4In summary, the conversation discusses the process of finding the maximum range of a projectile shot from an elevated angle on a hill. The conversation includes a discussion of the correct coordinate system to use, the equation for projectile motion, and the equation for the hill's slope. The final solution involves finding the maximum range by maximizing an equation with respect to the angle of elevation.
  • #1
Shackleford
1,656
2
http://i111.photobucket.com/albums/n149/camarolt4z28/14q.jpg?t=1284229871

Here is my work. My denominator is different than the book's answer for (a). I still don't know how to maximize alpha for (b) or maximize the range for (c).

http://i111.photobucket.com/albums/n149/camarolt4z28/141.jpg?t=1284229758
http://i111.photobucket.com/albums/n149/camarolt4z28/142.jpg?t=1284229758

http://i111.photobucket.com/albums/n149/camarolt4z28/14a.jpg?t=1284229759
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
If you're going to use the coordinate that you set, you'll have to watch out for correctly accounting for gravity. In both coordinate frames you will have a contribution from gravity. You actually even have your initial drawing of the coordinate system labeled incorrectly.
 
  • #3
Mindscrape said:
If you're going to use the coordinate that you set, you'll have to watch out for correctly accounting for gravity. In both coordinate frames you will have a contribution from gravity. You actually even have your initial drawing of the coordinate system labeled incorrectly.

I thought I did correctly account for gravity. Determining the effective gravity in this case is akin to dealing with a block on a plane.

What's wrong with the coordinate system? My understanding of the problem is a projectile with elevation alpha being shot up a hill of elevation. I suppose another interpretation could be a projectile shot with elevation alpha with respect to a hill of elevation beta.
 
  • #4
Select the co-ordinate system such that the direction OX i.e. X-axis along the surface of the inclined plane and Y-axis perpendicular to the surface.

Resolve vo into two components.
vox = vo*cos(α - β)
voy= vo*sin(α - β)
gx = g*sinβ and gy = g*cosβ

Τhe equation of the projectile motion is given by
y = vo*sin(α - β)*T - 1/2*g*cosβ*Τ^2

The time of flight T can be found by putting y = 0.

T = 2vo*sin(α - β)/g*cosβ

The horizontal distance x is given by
x = vo*cosα*T = 2vo^2*sin(α - β)*cosα/g*cosβ
Range d = x/cosβ = 2vo^2*sin(α - β)*cosα/g*cos^2(β)

Using the trigonometric identity 2sin(A)con(B) = sin(A+B) + sin(A-B), You can write

d = [vo^2/g*cos^2(β)]*[sin(2α - β) + sin(β)]

To find the angle of projection for maximum range,find derivative of d with respect to α and equate it to zero.

Substitute this angle in d to get the maximum range.
 
  • #5
Wait a second. I'm not sure why I have it backwards. Let me re-work it.
 
  • #6
I found x to be

x' = -(1/2) t^2 sin β + v0 cos (α - β)t.

I thought they wanted the x distance up the hill, or x', not the "original" x. That of course is simply x = v0 cos (α) t since there is no effective gravitational force in that direction. In x' coordinate transformation, there is an effective gravitational force along x' and y'.

Okay, I see what you did. To find d up the hill, you compared the x distance in the "original" system with d up the hill by cosine of beta determined by time T with y' = 0.

Well, from my d(d)/dα, I end up with cos(α - β) = 0.

d(d)/dα = [vo^2/g*cos^2(β)]*[2 cos(2α - β) + 0]

cos(2α - β) = 0

2α - β = n*(pi/2), where is n odd integer

α = β/2 + pi/4

I plugged in that angle, but I didn't get dmax.
 
Last edited:
  • #7
Range d = x/cosβ = 2vo^2*sin(α - β)*cosα/g*cos^2(β)

= 2vo^2*sin[π/4 +β/2 - β]*cos(π/4 +β/2)/g*(1 - sin^2β)

= 2vo^2*sin[π/4 - β/2]*cos(π/4 +β/2)/g*(1 - sin^2β)

= vo^2*[sinπ/2 - sinβ]/g*(1+sinβ)(1-sinβ)

= vo^2*/g*(1 + sinβ)
 
Last edited:
  • #8
I personally think this problem would be easier with a coordinate system in line the legs of the triangle, or the hill. I guess whatever is easiest, though you seemed to get in trouble with your gravitation earlier.

Of course, in the end you will get what rl.bhat posted.

Your condition is right. You should get

[tex]d_{max} = \frac{v_0^2}{g(1+sin \beta)}[/tex]

which makes sense because you know that it will go furthest when pointed directly up the hill, when beta is zero.
 
  • #9
Mindscrape said:
I personally think this problem would be easier with a coordinate system in line the legs of the triangle, or the hill. I guess whatever is easiest, though you seemed to get in trouble with your gravitation earlier.

Of course, in the end you will get what rl.bhat posted.

Your condition is right. You should get

[tex]d_{max} = \frac{v_0^2}{g(1+sin \beta)}[/tex]

which makes sense because you know that it will go furthest when pointed directly up the hill, when beta is zero.

I completely overlooked/forgot about the effective gravitation in the x' direction and switched my sines and cosines.

Initially, I plugged in the alpha-max in the last distance equation after the trigonometric identity was inserted. That of course yielded something different.
 
  • #10
Mindscrape said:
I personally think this problem would be easier with a coordinate system in line the legs of the triangle, or the hill. I guess whatever is easiest, though you seemed to get in trouble with your gravitation earlier.

Of course, in the end you will get what rl.bhat posted.

Your condition is right. You should get

[tex]d_{max} = \frac{v_0^2}{g(1+sin \beta)}[/tex]

which makes sense because you know that it will go furthest when pointed directly up the hill, when beta is zero.
Here is the simple method. Equation of the projectile motion is

y = x*tanα - 1/2*g*x^2/v^2cos^2α

Equation of the hill is given by

y = tanβ*x. Therefore

x*taνβ = x*tanα - 1/2*g*x^2/v^2cos^2α

So x = (tanα - tanβ)*2v^2*cos^2α/g = 2v^2cosα*sin(α - β)/g*cosβ.

And d = x/cosβ
 

FAQ: Projectile Motion: Launching Up a Hill

What is projectile motion?

Projectile motion is the motion of an object that is launched into the air and moves along a curved path under the influence of gravity.

How does launching up a hill affect projectile motion?

Launching up a hill affects projectile motion by changing the initial velocity and angle of the object, as well as the force of gravity acting on the object. This results in a different curved path compared to launching on flat ground.

What factors affect the trajectory of a projectile launched up a hill?

The trajectory of a projectile launched up a hill is affected by the initial velocity, angle of launch, and the height and slope of the hill. Air resistance and wind can also impact the trajectory.

How can we calculate the maximum height and range of a projectile launched up a hill?

The maximum height and range can be calculated using the equations for projectile motion, taking into account the initial velocity, launch angle, and the height and slope of the hill. These calculations can also be done using computer simulations.

In what real-life situations can we observe projectile motion when launching up a hill?

Projectile motion when launching up a hill can be observed in sports such as skiing, snowboarding, and skateboarding. It can also be seen in activities like launching a water balloon from a slingshot or throwing a ball up a hill.

Back
Top