Kinematic equations for projectile motion.

  • Thread starter Thread starter annalian
  • Start date Start date
  • Tags Tags
    height speed
Click For Summary

Homework Help Overview

The discussion revolves around kinematic equations related to projectile motion, specifically focusing on a particle thrown at an angle of π/6 radians. The original poster seeks to determine the initial speed and maximum height of the projectile based on given time intervals for ascent and descent.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the derivation of specific kinematic equations used in the problem, questioning the validity and source of these expressions. There is an exploration of the relationship between height at different times and the reasoning behind the equations applied.

Discussion Status

The conversation is ongoing, with participants seeking clarification on the origins of the equations presented. Some have offered insights into the reasoning behind certain expressions, while others express confusion and request further justification for the equations used.

Contextual Notes

There appears to be a lack of consensus on the correctness of the equations and assumptions made, with participants highlighting the need for clearer derivations and justifications for the expressions involved in the problem.

annalian
Messages
56
Reaction score
1

Homework Statement


The particle thrown under the angle pi/6 rad. with the horizontal direction achieves the height h for t1=10S when going up and t2=50S when going down. Find initial speed v0 and height h.

Homework Equations


h=v0y+gt^2/2

The Attempt at a Solution


As h1=h2
v0y*t1-gt1^2/2=gt2^2/2
v*0.5*10-10*100/2=10*2500/2
v0=2600.
hmax:
v^2-vy0^2=2gh
vy0=v0*sin30=2600*0.5=1300
h=0-1300^2=2*(-10)*h
h=84500m.
Am i right?
 
Physics news on Phys.org
annalian said:
v0y*t1-gt1^2/2=gt2^2/2
Where does this expression come from?
annalian said:
v^2-vy0^2=2gh
Where does this expression come from?

Just below the reply box is a link to a LaTeX guide which you may find helpful for writing formulae. For example ##h=v_{0y}t+\frac {1}{2}at^2##.
 
Ibix said:
Where does this expression come from?
Where does this expression come from?

Just below the reply box is a link to a LaTeX guide which you may find helpful for writing formulae. For example ##h=v_{0y}t+\frac {1}{2}at^2##.
Can you help me where to find the correct expressions for this exercise?
 
The expression you quoted in part 2 of your question seems sensible, and your reasoning that ##h_1=h_2## seems reasonable. However, I think you need to look at how you got from the equation in part 2 to the first expression I quoted. Post it here if you don't spot anything wrong.

You haven't provided any justification for the second expression at all. Where did it come from? It isn't immediately obvious that it's one of the kinematic formulae. I'm not saying it's wrong, just that I don't know where you got it from. I can guess, but it's easier if you tell me.
 
Ibix said:
The expression you quoted in part 2 of your question seems sensible, and your reasoning that ##h_1=h_2## seems reasonable. However, I think you need to look at how you got from the equation in part 2 to the first expression I quoted. Post it here if you don't spot anything wrong.

You haven't provided any justification for the second expression at all. Where did it come from? It isn't immediately obvious that it's one of the kinematic formulae. I'm not saying it's wrong, just that I don't know where you got it from. I can guess, but it's easier if you tell me.
v^2-vy0^2=2gh
As we know v^2-v0^2=2gh, but as we have two vectors of vo we should only use the one voy.
 
annalian said:
v^2-vy0^2=2gh
As we know v^2-v0^2=2gh, but as we have two vectors of vo we should only use the one voy.
Do we know that? Why? Neither of the expressions you've written makes sense to me, and you aren't explaining why you think they should.

Start with a statement of the law or principle or whatever you think you are applying to get those equations.
 

Similar threads

Replies
40
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 30 ·
2
Replies
30
Views
3K
Replies
13
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 14 ·
Replies
14
Views
5K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K