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A rock thrown... 
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#1
Sep1113, 08:09 AM

P: 43

A rock is tossed straight up with a speed of 20m/s. When it returns, it falls into a hole 10m deep.
A)What is the rock's velocity as it hits the bottom of the hole? So I drew the diagram and labelled it as I was taught to do, and then Googled the question to maybe see if I could get a pointer on how it works. The answer I saw that best fit the question was saying that the initial velocity is 0 m/s and that the final velocity is 20 m/s. I understand I need to use the kinematic equations to solve it. But if the question is asking what the final velocity is, why would I assume that the final velocity is 20 m/s? I would assume that the initial velocity is 20 m/s. Or would I assume that like they said, that the initial velocity is 0 m/s and the final velocity is 20m/s only to find time with a=ΔV/ΔT, using standard acceleration as 9.7m/s^{2}, and then go from there? I guess I'm just not understanding how we can have a "final velocity" if the question is asking for final velocity. 


#2
Sep1113, 08:14 AM

P: 43

Or is it trying to tell me that initial velocity should be 0 m/s and that at some point on it's journey upward it reaches 20 m/s, and that we wouldn't know the downward journey until we solved some equation for position at 10m?



#3
Sep1113, 08:20 AM

P: 1,320

The initial velocity is 20m/s .Final velocity is what needs to be calculated .The displacement between the final and initial position is given .
Do you know the three kinematics equations ? 


#4
Sep1113, 08:27 AM

P: 266

A rock thrown...
Also, it is helpful to break the problem up into sections, so the journey up would be one problem, where the initial velocity is 20 m/s and the final velocity is the velocity at its very highest point. Then the next part is its downward journey starting with its highest point and ending with it landing in the ground.



#5
Sep1113, 09:54 AM

P: 43

So I'm right to assume that initial velocity is 20m/s? The displacement would be 10m, acceleration would be 9.8m/s^{2}. So I could use the equation for velocity and displacement relation with uniform motion ((V_{f})^{2}=(V_{i})^{2}+2gΔx)? The reason I googled it was because I wasn't sure about initial position, but I guess it's always safe to assume initial position is 0m unless otherwise stated?



#6
Sep1113, 10:14 AM

P: 266

You're not assuming anything, the initial velocity is given. The rock has two parts to its motion, it is thrown UP and then it falls down. You need to find how high the rock travels upwards, ymax. And then from there its a simple free fall problem.
Consider this, what is the velocity at the peak of its motion? What is its acceleration at that point? 


#8
Sep1113, 10:21 AM

P: 43

Velocity at the rocks peak would be 0m/s and acceleration would be 9.8m/s/s.



#9
Sep1113, 10:28 AM

P: 266

Ok, so you have all the information to plug values into a formula. What formula has initial and final velocity, acceleration, initial and final position, final position is your unknown.



#10
Sep1113, 10:35 AM

P: 1,320




#11
Sep1113, 11:43 AM

P: 43

Ok. Thank you Tanya. Jesse, I would use the formula that I've already posted and broke up Δx into (y_{f}y_{i}), and then solve for y_{f}.



#12
Sep1113, 12:13 PM

P: 43

So I've solved for my final velocity and found it to be approximately 24.4m/s, is it acceptable to use "found" variables in future equation. For example this question has two parts, the first to find final velocity and the second to find t_{f}. Would it be acceptable to use final velocity I found in part A to find final time? I'm not really seeing any other way.
Edit: I should add, I remember when taking Chemistry that it wasn't usually a good idea to use variables found in previous parts of questions, just wondering if the same applies here. 


#13
Sep1113, 12:18 PM

P: 266

Staying consistent with variables is pretty much a required. Once the variables are defined then you need to stick with those otherwise it would be impossible to follow your work. That's true for any use of variables.



#14
Sep1113, 12:24 PM

P: 43

So, just to clarify, I'm able to use 24.4 m/s in future equations regardless if it was found by my own work?



#15
Sep1113, 03:00 PM

P: 266

I thought you meant variables as in "v" for velocity.
Generally you'll have multi part questions and most of the time you need values from previous parts to solve subsequent parts. So yes, if you found that answer in the same problem you probably have to use it again. Just be sure that it is indeed the value that you need. 


#16
Sep1113, 08:58 PM

P: 43

Thank you muchly. I've concluded that question and solved for all answers, and there was much rejoicing.



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