Physics Help Forum [SOLVED] Classical mechanic problem/energy laws help

 Aug 15th 2008, 01:45 PM #1 Senior Member     Join Date: Apr 2008 Posts: 815 [SOLVED] Classical mechanic problem/energy laws help I didn't try my best to solve it, but really I'm not so sure and never done such an exercise so I'd like to see how to solve those types of exercises. A projectile is fired horizontally by a canon situated a platform at an height of 44 m. It is fired with an initial velocity of 244 m/s. Suppose the ground is perfectly plane and horizontal. Using conservative energy laws find out : 1) The magnitude of the vertical component of the velocity of the projectile when it reaches the ground. 2)Do the same calculus in the case the projectile is falling off from 44 m. My attempt : I know that the modulus of the velocity won't change in the 2 cases, but I'll have to show it. My real attempt : I know that $\displaystyle E=\frac{mv^2}{2}+mgz$ when the weight (I call it "$\displaystyle P$" in equations) is $\displaystyle \vec{P}=-m \vec{g} z$. It means I use a z-axis instead of the commonly used y-axis. Having said that, the horizontal and vertical components confuses me when I must use the law of conservation of energy. I feel very unsure. I start saying that the energy of the system is the same at the beginning and at the final. Mathematically I wrote $\displaystyle \frac{mv_i^2}{2}+mgz_i=\frac{mv_f^2}{2}+mgz_f$. The mass cancels out, so it's $\displaystyle \Leftrightarrow$ to say $\displaystyle \frac{v_i^2}{2}+gz_i=\frac{v_f^2}{2}+gz_f$. I believe I'm looking for $\displaystyle v_f$. Oh, I just got an image in my head. Once I get $\displaystyle v_f$, I get the modulus of $\displaystyle v_f$, right? It means that I just have to play with Pythagoras laws in order to get $\displaystyle v_{f,z}$ because I already have $\displaystyle v_x$ and because it is a constant I have $\displaystyle v_{x,f}$. So let's try it out. $\displaystyle v_f^2=v_i^2+2gz_i-2gz_f$ So $\displaystyle v_f=\sqrt{v_i^2+2gz_i-2gz_f}$. As $\displaystyle z_f=0$, it simplifies at $\displaystyle v_f=\sqrt{v_i^2+2gz_i}$ Which is worth 61260.8 m/s. Hence $\displaystyle v_f$ is approximately equal to 247.5091917 m/s. Which seems possible to me, but below what I expected. So Pythagoras says $\displaystyle v_{z,f}$ is approximately equal to 41.53... well, doesn't make sens. What did I wrong? __________________ Isaac If the problem is too hard just let the Universe solve it. Last edited by arbolis; Aug 15th 2008 at 05:19 PM.
Aug 16th 2008, 01:18 AM   #2
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Hello Arbolis !
 Originally Posted by arbolis So Pythagoras says $\displaystyle v_{z,f}$ is approximately equal to 41.53... well, doesn't make sens. What did I wrong?
Until this sentence, what you've done is correct. The mistake I've found is that, using the Pythagorean theorem, $\displaystyle v_f^2 = v_{z,f}^2+v_i^2$ so

$\displaystyle |v_{z,f}|=\sqrt{v_f^2-v_i^2}=\sqrt{v_i^2+2gz_i-2gz_f-v_i^2}=\sqrt{2gz_i}=29\text{ m/s}\neq 42\text{ m/s}$

Does this result make more sense then the one you've found ? If your answer to this question is "No", try to answer question #2 : "Do the same calculus in the case the projectile is falling off from 44 m."

 Aug 16th 2008, 10:07 AM #3 Senior Member     Join Date: Apr 2008 Posts: 815 Oh nice... thanks a lot. I didn't thought about back-subing the $\displaystyle v_f^2$. But I don't understand why what I did doesn't work. From $\displaystyle |v_{z,f}|=\sqrt{v_f^2-v_i^2}$, I just replaced $\displaystyle v_f^2$ by $\displaystyle 147.5$ and $\displaystyle v_i$ by $\displaystyle 244$. And it gives $\displaystyle 41.5$ m/s as a result. Oops! $\displaystyle v_f^2=247.5$! Now it gives about $\displaystyle 38.4$ m/s which is still not good. What did I wrong? __________________ Isaac If the problem is too hard just let the Universe solve it.
Aug 16th 2008, 12:03 PM   #4
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 Originally Posted by arbolis Oops! $\displaystyle v_f^2=247.5$!
It seems you've inverted two digits : $\displaystyle v_f=\sqrt{244^2+2\times 9.81 \times 44}=24{\color{red}5}.{\color{red}7}$ so $\displaystyle |v_{z,f}|=\sqrt{245.7^2-244^2}=29\text{ m/s}$.

Aug 16th 2008, 01:40 PM   #5
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Join Date: Apr 2008
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 It seems you've inverted two digits
Oh... thanks a lot. It made such a big difference!
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