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A simple problem really.

Part 1 Suppose a ball ($20. ~kg$) was shot straight up with an initial velocity of $+50~m/s$.

a. Assuming that all of the ball's initial $E_k$ was transformed into $E_g$, what is the maximum height the ball could reach?

My work: $$E_g=E_k$$ $$mgh = 0.5mv^2$$ $$h=v^2/2g$$ $$50^2/2(9.8) = h$$ $$h=128m$$

Assuming this is correct let's move on to Part 2.

Part 2 Suppose that 20% of the ball's initial $E_k$ was lost due to friction with the air (air resistance). What is the maximum height the ball could reach?

I thought of this initial equation first:

$$E_g = E_k + E_f$$ Where $E_f$ = Air Resistance

Apparently this equation is not right and the right one is

$$E_k = E_g + E_f$$

I thought the equation above makes no sense at all, and if someone could intuitively explain and conceptualize why this is true and why the statement above makes sense, I'd be happy.

Thanks guys. Hopefully formatting is OK? (I'm a Chem /ce{} format-er).

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  • $\begingroup$ I think that in Part (2) the use of Eg, Ek, and Ef to set up an equation is a bit confusing. I would simply note that if 20% of the ball's initial Ek is lost due to friction, then you can can immediately write down Eg=(0.8)Ek for your first equation in part (1) (that is, if 20% of Ek is lost to friction then only 0.8Ek is left to contribute to the potential gravitational energy of the object at max height). After that follow the same path used in part (1) to get the answer for the height h with friction. $\endgroup$ – Samuel Weir Nov 4 '15 at 3:07
  • $\begingroup$ @SamuelWeir You could. That's how I would've initially done it, but, how can it be done this way? $\endgroup$ – Asker123 Nov 4 '15 at 3:10
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    $\begingroup$ OK, then look at it this way: The projectile has some initial kinetic energy Ek. An amount Ef of that energy is lost to friction. That means that only Ek-Ef energy remains to contribute to the gravitational potential energy of the projectile at max height. Therefore Eg=Ek-Ef, which is equivalent to the 2nd of your equations in part (2). $\endgroup$ – Samuel Weir Nov 4 '15 at 3:13
  • $\begingroup$ Please give your question a meaningful title $\endgroup$ – EnergyNumbers Nov 4 '15 at 7:44
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In the first part you wrote $E_{k}=E_{g}$ because kinetic energy is fully converted into potential energy. But in the second part, some of the initial kinetic energy $(E_{f})$ lost due to friction and part of energy left is $E_{k}-E_{f}$ . Only this part is converted to potential energy $E_{g}$ . Thus,

$E_{g}=E_{k}-E_{f}$ and this simplified as

$E_{k}=E_{g}+E_{f}$

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The initial kinetic energy $E_k$ gets partly dissipated as friction, $E_f$, and partly converted to gravitational potential energy, $E_g$. The sum of these two must equal the original energy input, so

$$E_k = E_f + E_g$$

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