Summoning all math & physics newbies:
I thought of an interesting problem to work on if you haven't learned it already. Let's work on it together, ruling out ideas and coming to a conclusion without cheating.
If you throw a ball up in the air at any given velocity, what will be the height of the ball the moment it begins to return to the Earth?
My lead so far is that two variables are responsible for this: the initial velocity when the ball is first released, and the gravitational acceleration constant since all objects accelerate the same in any given gravitational field regardless of mass.
I know the time it takes to reach the top should not be a variable in the equation because all initial velocities will behave the same way, so that if you throw two different weights up, they will reach a height proportionally.
So I have now g and V_i. I know that as the velocity increases, the height increases, so I can expect that in the numerator, and I know as the gravitational acceleration constant increases, the height decreases, so I can expect to see that in the numerator.
v_i / g
But now I'm stuck! I know I'm not taking into account when the velocity is equal to 0 in this expression.
Hmmmm... any thoughts? Don't spoil the fun if you know the answer already! I like to think as part of a teamwork effort. Thank you!
>>8043584
Oops! EDIT:
>the height decreases, so I can expect to see that in the denominator.
If you don't know Vi you won't know how high the ball gets, no way. Acceleration downwards by gravity acts on the speed of the ball. If the ball is going 100 m/s it will take longer for acceleration downwards to bring it to a complete stop than if it was going 50m/s
>>8043589
But this problem is assuming you do know Vi. If you throw a ball into the air (a predetermined initial velocity), what will its height be when it reaches the top?
>>8043589
Sorry ... I see what you mean now
>>8043589
Obviously. He's asking for the height of the ball in terms of V_i and g.
>>8043589
Oh sorry, I see what you are saying now. Yes. I guess we need another variable of what the velocity is in any given moment when time is equal to 0. And when /that/ is equal to 0, then we've reached the top. I wonder how to incorporate this into the formula.
>>8043601
when the change of time that has elapsed is equal to 0**
I'm hitting the drawing board. I'll be back in a bit on further leads!
The acceleration due to gravity is F/m, and the force due to gravity is GMm/r^2. So a(r)=GM/r^2. We want to find the moment when the kinetic energy is zero and the potential energy is at it's max. We know that the kinetic energy is initially at a max of (1/2)mv^2, and by conservation of energy the max potential energy is m*a(r)*r. So we know that (1/2)mv^2=m*a(r)*r, and the mass of the object cancels out. substituting in for a(r) and we get (1/2)v^2=r*GM/r^2. Solving for r we get r=2GM/v^2. G is the gravitational constant and M is the mass of the earth, in case that was unclear.
>>8043597
Isn't it just -Vi^2/2g from the newtons equations
>>8043615
Sorry, I made a mistake. r is the distance from the center of the earth, so the max potential energy should be m*a(r)*(r-R) where R is the radius of the earth. This means that (1/2)v^2=(r-R)GM/r^2.
So (1/2)v^2=(GM/r)-(RGM/r^2). Multiplying both sides by r^2 gives us the quadratic equation [math] \frac{1}{2} v^2 r^2 -GMr+RGM=0 [/math]
[eqn] r= \frac {GM \pm \sqrt{ (GM)^2 -4 \frac{1}{2} v^2 RGM} }{v^2} [/eqn]
The quantity we're interested in is actually (r-R).
>>8043615
Hmm. I don't know much about kinetic and potential energy. I don't see how that would be relevant here given that the height seems to only depend on the initial velocity and gravitational constant (which can be anything really; it just depends on the body so technically you can call it a variable) but it could very well be.
>>8043679
Can you explain your reasoning of why you think that is relevant? I'm interested. Maybe I can glean some insight here.
>>8043683
Just think about it. At the ground the potential energy is zero, and you throw the ball with an initial velocity. As the ball goes up, gravity slows it's velocity down until it comes to a stop and turns back. So we can say that the kinetic energy of the ball is at a max at the ground. Just to recap: at the ground, the potential energy is zero and the kinetic energy is max.
Now think about the maximum height of the ball. As I said, it comes to a stop and turns around. So since it's velocity is zero at this point, the kinetic energy is zero. And because it's at it's maximum height, it's potential energy is max. In other words, the kinetic energy is all transferred into potential energy on the way up, and the potential energy is converted back into kinetic energy on the way down.
We can use this in the problem because we know that energy is always conserved, so the max in potential energy and the max in kinetic energy will be the same. We know the potential energy is m*g*h (here I use a(r) instead of g to be a bit more general) and we know kinetic energy is (1/2)*m*v^2. These can be set equal to each other when v=v_i and when h=(r-R). Using potential and kinetic energy is how the kinematic equations are derived, so it's quite relevant to use them for this kind of problem.
>>8043683
This is a conservation of energy problem. You need to realize when the kinetic energy is zero, the potential energy is mgh, where h is the highest height of the ball. When you initially hurl the ball up in the air, it has a certain kinetic energy, mv^2/2, where v is velocity. Since energy is conserved, you can set m(v_i)^2/2 = mgh, and solve for h = (v_i)^2/(2mg).
>>8043707
You're assuming the ball is going significantly into the atmosphere, here. For cases like throwing a ball, you can just use 9.8 m/s^2
>>8043726
Whoops, v_i^2/(2g), typo, am high
>>8043728
I didn't assume anything. My equation works for balls thrown into low or high atmospheres of any mass M with radius R. You're the one making the assumption that acceleration due to gravity is 9.8 m/s^2
