Kinematic equations for projectile motion.

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The discussion focuses on solving a projectile motion problem involving a particle thrown at an angle of pi/6 radians. The key equations used include h = v0y*t + (1/2)gt^2, where h is the height, v0y is the initial vertical speed, and g is the acceleration due to gravity. Participants question the derivation and application of kinematic equations, particularly the transition from one expression to another. There is a consensus that the reasoning behind the equations needs clarification, especially regarding the use of initial vertical velocity. The thread emphasizes the importance of understanding the principles behind the kinematic formulas used in projectile motion calculations.
annalian
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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?
 
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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.
 
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