Can You Prove Conservation of Energy in Vertical Stone Throwing?

AI Thread Summary
The discussion focuses on proving the conservation of energy in the context of a stone thrown vertically upward with air drag acting on it. The maximum height reached by the stone is derived as h = v(i)^2 / 2g(1+f/w), and the speed just before impact is v = v(i) (w-f/w+f)^1/2. Participants emphasize the need to apply the conservation of energy equation and Newton's second law to relate forces and acceleration. There is a detailed exploration of how to account for the forces acting on the stone, including weight and air drag, while ensuring proper sign conventions. Ultimately, the contributors clarify the relationship between acceleration and the equations of motion, leading to a better understanding of the problem.
jenavira
Messages
10
Reaction score
0
This is driving me nuts.

A stone with weight w is thrown vertically upward into the air from ground level with initial speed v(i). If a constant force f due to air drag acts on the stone throughout its flight, (a) show that the maximum height reached by the stone is

h = v(i)^2 / 2g(1+f/w)

and (b) show that the stone's speed just before impact with the ground is

v = v(i) (w-f/w+f)^1/2


I've been working through these for ages and I'm just lost. I know I need the conservation of energy equation 1/2mv^2 = mgh and probably the kinetic motion equation 2a(delta x) = v(f)^2 - v(i)^2, but I'm just not sure what to do with them.
 
Physics news on Phys.org
find the stone's acceleration using Newton's 2nd law.
 
Huh. I'm getting closer. But I still have no idea how I'm supposed to relate the conservation of energy and kinetic motion equations. (It's probably something really obvious, and I'll feel like an idiot for not realizing it sooner, but I've been staring at these for so long...)
 
what is the acceleration that you have got ?
 
Well, F = ma so a = F/m...and the force in question would be the initial velocity. But I'm not at all sure how that helps...
 
If upward direction is taken to be positve than using Newton's 2nd law you get
-f+(-W)=ma but since W=mg than m=W/g
then you get a = -g(1+f/w) and using this constant acceleration substistute into
V^2=u^2+2as to get the max height. becareful with the sign.
 
...right. I'm going to sound like an idiot, but could you go through that accelleration in a bit more detail? I'm still a bit confused.
 
\Sigma\vec{F}=m\vec{a} ,right /
two downward forces which are the stone weight and the drag force act on the stone when it is moving upward ,right ?
if the upward direction is chosen to be positive, then downward direction must be negative. then we have -W for the stone weight and -f for the drag force. negative signs indicate the direction and W and f are their magnitudes.
you know that the magnitude of the stone weight is given by W= mg W, m and g are all positive because i am talking about magnitudes.
then substitute into the equation above , you will get what i get.
 
I think I've managed it now. Thanks a lot. :)
 
Back
Top