# Kinematic equations for projectile motion.

• annalian
In summary: 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##.In summary, the equation for the height h of a particle thrown under the angle pi/6 rad. with the horizontal direction is h=v0y+gt^2/2. The equation for the height h of a particle thrown under the angle pi/6 rad. with the horizontal direction is h=v0y+gt^2/2.
annalian

## 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.

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?

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.

## 1. What is the formula for finding the speed v0 and height?

The formula for finding the speed v0 and height is v0 = √(2gh), where v0 is the initial velocity, g is the acceleration due to gravity (9.8 m/s²), and h is the height.

## 2. How do I find the initial velocity using this formula?

To find the initial velocity using the formula v0 = √(2gh), you need to know the acceleration due to gravity and the height. Plug these values into the formula and solve for v0.

## 3. Can this formula be used for any object thrown or dropped from a height?

Yes, this formula can be used for any object thrown or dropped from a height as long as the acceleration due to gravity remains constant and there is no air resistance or other external forces acting on the object.

## 4. How accurate is this formula in real-world situations?

This formula provides an accurate estimate of the initial velocity and height for objects dropped or thrown in a vacuum. In real-world situations, there may be other factors such as air resistance that can affect the accuracy of the results.

## 5. Can this formula be used to find the speed and height of an object at any point during its motion?

No, this formula can only be used to find the initial velocity and height of an object. To find the speed and height at any point during its motion, you would need to use other equations that take into account the object's acceleration and time.

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