Problem about projectile motion

Click For Summary

Homework Help Overview

The discussion revolves around a problem related to projectile motion, specifically involving a parabolic trajectory and its interaction with a hemisphere, as presented in a textbook by Serway. Participants are examining the conditions under which a projectile may or may not intersect with the hemisphere based on given equations and numerical values.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants are questioning the validity of certain equations and their implications regarding the trajectory of the projectile. There is a focus on plugging in values to test the equations and understanding the geometric relationship between the parabolic path and the hemisphere.

Discussion Status

The discussion is active, with participants providing insights and suggestions for further exploration. Some guidance has been offered regarding numerical calculations and the interpretation of the equations, but there is no explicit consensus on the solution yet.

Contextual Notes

Participants are working with specific equations from their notes and the textbook, and there is a mention of the need for numerical calculations to validate their assumptions. The original poster has provided their own solutions and referenced a solution manual, indicating a structured approach to the problem.

issacnewton
Messages
1,035
Reaction score
37
hi

i have this problem from serway's book. i have posted problem and its solution (from solution manual which i have) in 1.doc my own solution is in r.pdf. i have explained my problem there.
please read it and try to answer

Newton
 

Attachments

Physics news on Phys.org
Hi IssacNewton,

IssacNewton said:
hi

i have this problem from serway's book. i have posted problem and its solution (from solution manual which i have) in 1.doc my own solution is in r.pdf. i have explained my problem there.
please read it and try to answer

Newton

I have not checked all your equations, but I don't believe eq. 15 from your notes shows that the rock never touches the hemisphere. Have you tried plugging in values for x between 0 and R, and finding the position of the rock?

Alternatively, you could write down the equation of a circle, and find where the parabola intersects the circle for x between 0 and R.
 
alphysicist said:
Hi IssacNewton,



I have not checked all your equations, but I don't believe eq. 15 from your notes shows that the rock never touches the hemisphere. Have you tried plugging in values for x between 0 and R, and finding the position of the rock?

hi

no i have not done the plugging between the endpoints. but at x=0, y is R and at y=0, x is 1.21R, so when the ball lands on the ground it lands beyond the hemisphere, also eq 15 is a parabola. if ball ever touches the hemisphere, how can we get that at y=0, x is 1.21R.
i have known the other approach shown by you, which is also there in the author's solution
posted by me. i am just trying to understand my solution.

thanks
 
IssacNewton said:
hi

no i have not done the plugging between the endpoints. but at x=0, y is R and at y=0, x is 1.21R, so when the ball lands on the ground it lands beyond the hemisphere, also eq 15 is a parabola. if ball ever touches the hemisphere, how can we get that at y=0, x is 1.21R.
i have known the other approach shown by you, which is also there in the author's solution
posted by me. i am just trying to understand my solution.

thanks

Let's consider your first case, which gave your Eq. 10. The problem is not that the path is touching the hemisphere at y=0, it is that the entire parabolic path up to that point is inside the rock.

So if you'll calculate some sample points for Eq. 15, I believe you'll find that although it ends up well past the hemisphere, the majority of that trajectory is also inside the rock. For example, set R=1, x=0.5, and find y; then use that to calculate x^2 + y^2. If that result is less than R^2=1, then that part of the path is inside the rock and the path is not viable.

If you think about how the points of intersection between the parabola and hemisphere behave as the initial speed is increased, I think you'll see why the book's answer is the lowest possible speed.
 
alphysicist said:
Let's consider your first case, which gave your Eq. 10. The problem is not that the path is touching the hemisphere at y=0, it is that the entire parabolic path up to that point is inside the rock.

So if you'll calculate some sample points for Eq. 15, I believe you'll find that although it ends up well past the hemisphere, the majority of that trajectory is also inside the rock. For example, set R=1, x=0.5, and find y; then use that to calculate x^2 + y^2. If that result is less than R^2=1, then that part of the path is inside the rock and the path is not viable.

If you think about how the points of intersection between the parabola and hemisphere behave as the initial speed is increased, I think you'll see why the book's answer is the lowest possible speed.

Eureka...Wow.. Al ... I never considered the possibility that eq 10 could describe a parabola lying entirely inside the hemisphere. i never did numerical calculations... and the eq 15, its so close to the answer but at x=0.2 R , x^2+y^2 = 0.98 R^2, so its still inside the hemisphere. so even that is not the solution. even if i was wrong, my struggle gave me great insight in the problem, which is necessary in problem solving... now i will look at the solution given by serway...

thanks
 
IssacNewton said:
Eureka...Wow.. Al ... I never considered the possibility that eq 10 could describe a parabola lying entirely inside the hemisphere. i never did numerical calculations... and the eq 15, its so close to the answer but at x=0.2 R , x^2+y^2 = 0.98 R^2, so its still inside the hemisphere. so even that is not the solution. even if i was wrong, my struggle gave me great insight in the problem, which is necessary in problem solving... now i will look at the solution given by serway...

thanks

Sure, glad to help!
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 2 ·
Replies
2
Views
806
Replies
9
Views
2K
Replies
40
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 18 ·
Replies
18
Views
2K
  • · Replies 38 ·
2
Replies
38
Views
7K
Replies
19
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K