Conservation of energy problem (projectile with air drag)

AI Thread Summary
The discussion focuses on a physics problem involving a stone thrown vertically upward with air drag. The maximum height formula derived is h = v_0^2 / (2g(1+f/w)), and the challenge lies in finding the stone's speed just before impact, represented as v = v_0 * ((w-f)/(w+f))^1/2. The user struggles with the calculations for part (b), particularly in equating work done by air drag to the change in kinetic energy. They receive feedback indicating that the error lies in the last line of their calculations. The conversation emphasizes the importance of correctly applying energy conservation principles in the presence of air resistance.
azure kitsune
Messages
63
Reaction score
0

Homework Statement



A stone with weight w is thrown vertically upward into the air from ground level with initial speed v0. 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 = \frac{v_0^2}{2g(1+f/w)}

(b) Show that the stone's speed is

v = v_0 \left( \frac{w-f}{w+f} \right) ^ {1/2}

just before impact with the ground.

Homework Equations



work done by an external force = change in energy

The Attempt at a Solution



I have no trouble with part (a). I need help with (b)

\begin{align*}<br /> W_{air} &amp;= (-f) * (2h) \\<br /> &amp;= -2fh \\<br /> &amp;= -2f\frac{v_0^2}{2g(1+f/w)} \\<br /> &amp; = -2f\frac{v_0^2 w}{2g(w+f)} <br /> \end{align*}

W_{air} = \Delta E = \Delta K = \frac{1}{2}m(v^2 - v_0^2) = \frac{w}{2g}(v^2 - v_0^2)

We can set the two expressions equal

\begin{align*}<br /> -2f\frac{v_0^2 w}{2g(w+f)} &amp;= \frac{w}{2g}(v^2 - v_0^2) \\<br /> -2f\frac{v_0^2 }{(w+f)} &amp;= v^2 - v_0^2 \\ <br /> v^2 &amp;= v_0^2 + -2f\frac{v_0^2 }{w+f} \\<br /> &amp;= v_0^2 \left( \frac{1-2f}{w+f} \right)<br /> \end{align*}

But this is wrong. Can anyone tell me where I messed up?
 
Physics news on Phys.org
azure kitsune said:
\begin{align*}<br /> v^2 &amp;= v_0^2 + -2f\frac{v_0^2 }{w+f} \\<br /> &amp;= v_0^2 \left( \frac{1-2f}{w+f} \right)<br /> \end{align*}

But this is wrong. Can anyone tell me where I messed up?

Hi azure kitsune! :smile:

Only the last line is wrong. :cry:
 
Waaahhhh! *feels so stupid!*

Thanks tiny-tim. :smile:
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top