Projectile problem proof

In summary, the conversation discusses finding the distance traveled by a projectile fired up an incline with an initial speed and angle with respect to the horizontal. It is shown that the distance can be expressed as d = 2*vi^2*cosθi*sin(θi-φ) / g*cos(φ)^2 and the maximum value of d and the corresponding angle θi can be determined using the usual equations for x and y, by rotating the coordinate system, and by using trigonometric identities. The conversation also delves into finding the time and coordinates of the landing point on the incline.
  • #1
golbez987
3
0
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

(b.) For what value of θi is d a maximum, and what is the maximum value?


I was given a hint to get d into the projectile equations by expressing x(final) and y(final) in terms of d and φ.

So far I haven't been trying to use that hint, but I can't figure out how...

Any help would be highly appreciated.
 
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  • #2
Rotate the x-axis so that it's parallel with the incline
 
  • #3
golbez987 said:
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

(b.) For what value of θi is d a maximum, and what is the maximum value?


I was given a hint to get d into the projectile equations by expressing x(final) and y(final) in terms of d and φ.

So far I haven't been trying to use that hint, but I can't figure out how...

Any help would be highly appreciated.

Write the usual equations for x and y ( [itex] x= v_i cos (\theta_i) t, y= v_i sin ( \theta_i) t - 1/2 g t^2 [/itex]).

Now, simply use [itex] x= d cos (\phi) [/itex] and [itex] y = d sin ( \phi) [/itex]. (this is just saying that the coordinates of the landing point are up at a distance d along the inclined plane).

Patrick
 
  • #4
That wouldn't mess up the acceleration on the y-axis? I was thinking that if I rotated my coordinate system the acceleration due to gravity would change direction in some way.
 
  • #5
golbez987 said:
That wouldn't mess up the acceleration on the y-axis? I was thinking that if I rotated my coordinate system the acceleration due to gravity would change direction in some way.

Yes, if you rotate the coordinate system, you have to watch for gravity. In the new coordinate system, you would have both an a_x and and an a_y due to gravity. It's not too hard, but it's probably easier to do it the way I describe in my post.

Patrick
 
  • #6
Ok. Thanks, that has already been a big help.
 
  • #7
That's right. [tex]g_x=g\sin\theta[/tex] and [tex]g_y=g\cos\theta[/tex]
 
  • #8
phucnv87 said:
That's right. [tex]g_x=g\sin\theta[/tex] and [tex]g_y=g\cos\theta[/tex]

It depends on the side of the inclined plane. If the plane is at the right
(as is usually used...I mean of the form "<"), then [itex] a_x= -g \cos (\phi) [/itex]. And it should be a [itex] \phi [/itex], to be aligned with the plane.

Pat
 
  • #9
It depends on the direction of the coordinate you choose to be positive
 
  • #10
phucnv87 said:
It depends on the direction of the coordinate you choose to be positive

LOL! You are right ;-)
 
  • #11
I've been working on this same problem for quite some time but I'm still stuck.

Following Patrick's method, i used "the usual x and y equations" and plugged in x = d(cos(phi)) and y = d(sin(phi)) into them.

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

after this point, I'm not sure how to determine the time (t) that i should plug into the equation.

also, once i find t, am i supposed to use the pythagorean theorem to find what point the projectile hits the incline?

thanks in advance
 
Last edited:
  • #12
Epiphone said:
I've been working on this same problem for quite some time but I'm still stuck.

Following Patrick's method, i used "the usual x and y equations" and plugged in x = d(cos(phi)) and y = d(sin(phi)) into them.

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

after this point, I'm not sure how to determine the time (t) that i should plug into the equation.

also, once i find t, am i supposed to use the pythagorean theorem to find what point the projectile hits the incline?

thanks in advance


Just isolate the time from the first equation and plug in the second equation. then isolate the distance d and you have your final answer: d expressed in terms of the angles phi and theta and the initial speed. To get the form quoted in the original post, you will ahve to play with trigh identities.

Patrick
 
  • #13
Epiphone said:
I've been working on this same problem for quite some time but I'm still stuck.

Following Patrick's method, i used "the usual x and y equations" and plugged in x = d(cos(phi)) and y = d(sin(phi)) into them.

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

after this point, I'm not sure how to determine the time (t) that i should plug into the equation.

also, once i find t, am i supposed to use the pythagorean theorem to find what point the projectile hits the incline?

thanks in advance


Just isolate the time from the first equation and plug in the second equation. then isolate the distance d and you have your final answer: d expressed in terms of the angles phi and theta and the initial speed. To get the form quoted in the original post, you will have to play with trigh identities. I haven`t double checked the equation quoted there but if there is no typo it should work

Patrick
 
  • #14
nrqed said:
Just isolate the time from the first equation and plug in the second equation. then isolate the distance d and you have your final answer: d expressed in terms of the angles phi and theta and the initial speed. To get the form quoted in the original post, you will have to play with trigh identities. I haven`t double checked the equation quoted there but if there is no typo it should work

Patrick

if i isolate t from the first equation, i get t = vicos(θi)/dcos(phi)
then i take this t and plug it into the 2nd equation,

dsin(phi) = visin(θi)t - .5gt^2

however, once i plug in my t value, i get a very complicated equation (especially because of the t^2) and i don't know how to isolate d from it. I'm not sure if i am quite following what you are saying.
 

1. What is a projectile problem proof?

A projectile problem proof is a mathematical method used to solve problems related to the motion of objects that are under the influence of gravity, such as a ball thrown into the air or a rocket launched into space.

2. How does a projectile problem proof work?

A projectile problem proof uses equations derived from the laws of motion and gravity to calculate the position, velocity, and acceleration of an object at different points of its trajectory.

3. What are the key components of a projectile problem proof?

The key components of a projectile problem proof include the initial velocity of the object, the angle at which it is launched, the acceleration due to gravity, and the time elapsed during the motion.

4. Can a projectile problem proof be used for any type of object?

Yes, a projectile problem proof can be applied to any object that is under the influence of gravity, as long as the initial conditions and the force of gravity are known.

5. What are some real-life applications of a projectile problem proof?

Some real-life applications of a projectile problem proof include predicting the trajectory of a missile, analyzing the motion of a baseball in a game, and understanding the flight path of a satellite in orbit.

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