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 Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum Jul 16th 2017, 04:36 PM #1 Junior Member   Join Date: Jun 2017 Posts: 15 Vertical displacement Hi, Currently working on a problem and I'm very lost, no detail in the lesson text on how to do it. Any help is appreciated. A spring is launched at 2.3m/s at an angle of 78 degrees to the ground. What is the maximum height reached by the spring? First I find the vertical velocity: sin78 = V1v / 2.3m/s 0.978 = V1v / 2.3m/s 2.2494 = V1v From here, I try to find the time because I believe that is the next step, but I'm not sure if I'm doing it right. I tried this: t = v/a t = 2.2494/9.8 t = 0.2295 seconds Assuming that's correct, I move on to what I think is the right equation: dv = V1v (t) + 0.5at^2 dv = 2.2494m/s(0.2295s) + 0.5(-9.8m/s^2)(0.2295)^2 dv = 0.51624 - 4.9(0.05267) dv = 0.51624 - 0.25808 dv = 0.25816m This answer just doesn't seem right to me, please let me know what I did right/wrong and what I should be doing differently. Thanks very much.   Jul 16th 2017, 08:03 PM   #2
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 Originally Posted by jpompey Hi, Currently working on a problem and I'm very lost, no detail in the lesson text on how to do it. Any help is appreciated. A spring is launched at 2.3m/s at an angle of 78 degrees to the ground. What is the maximum height reached by the spring? First I find the vertical velocity: sin78 = V1v / 2.3m/s 0.978 = V1v / 2.3m/s 2.2494 = V1v
Yes, that's correct.

 From here, I try to find the time because I believe that is the next step, but I'm not sure if I'm doing it right.
That is one way.

 I tried this: t = v/a t = 2.2494/9.8 t = 0.2295 seconds
That is the time the speed drops from 2.2494 to 0 at deceleration -9.8 m per second squared. Another way to do this is to write the equation for height:
s= (-g/2)t^2+ v0t= (-9.8/2)t^2+ 2.2494t= -4.9t^2+ 2.2494t.
The maximum height occurs when the derivative is 0: -9.8t+ 2.2494= 0. Of course, that is the same as saying that the vertical velocity at the top of the arc is 0 so that t= 2.2494/9.8 as you have.

 Assuming that's correct, I move on to what I think is the right equation: dv = V1v (t) + 0.5at^2 dv = 2.2494m/s(0.2295s) + 0.5(-9.8m/s^2)(0.2295)^2 dv = 0.51624 - 4.9(0.05267) dv = 0.51624 - 0.25808 dv = 0.25816m This answer just doesn't seem right to me, please let me know what I did right/wrong and what I should be doing differently. Thanks very much.
Looks good to me.   Jul 16th 2017, 08:13 PM #3 Junior Member   Join Date: Jun 2017 Posts: 15 thanks very much!    Jul 17th 2017, 05:51 AM #4 Senior Member   Join Date: Aug 2008 Posts: 113 In general, max height above launch position occurs when $v_y = 0$ $0=v_{y} = v_{y0} - gt \implies t=\dfrac{v_{y0}}{g}$ $\Delta y = v_{y0} \cdot t - \dfrac{1}{2}gt^2$ $h_{max} = v_{y0} \cdot \dfrac{v_{y0}}{g} - \dfrac{1}{2} g \cdot \dfrac{v_{y0}^2}{g^2}$ $h_{max} = \dfrac{v_{y0}^2}{g} - \dfrac{v_{y0}^2}{2g}$ $h_{max} = \dfrac{v_{y0}^2}{2g}$ $h_{max} = \dfrac{(v_0 \sin{\theta})^2}{2g}$ jpompey likes this.   Jul 17th 2017, 06:39 AM   #5
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 Originally Posted by jpompey t = v/a
If I may make note of a nitpicky point. It was pointed out by HOI above but I'd like to stress this point further: The equation for v (t) when a = constant is v = at + v_0. Since you're setting v = 0 so you should get t = -v_0/a. You got t = v_0/a. Then substitute a = -9.8m/s^2 to get your answer which is correct. I' m making this comment to stress the importance of a precisely correct equation and getting the sign correct. All too many time the mistake in a derivation comes from having the sign being wrong somewhere.

 Originally Posted by jpompey dv = 0.25816m This answer just doesn't seem right to me, ..
It seemed quite wrong to me too and I there is no error in your derivation.

Then I realized why it seemed wrong to me. I had the impression that 2.25m/s was a large speed. I was mistaken. That's because I'm not used to thinking in terms of m/s. I live in the USA and as such typically think in terms of mph. 2.25 m/s = 5 mph which is pretty slow. Given that, I'm comfortable with your answer. Note that in all my physics calculations over the last so many years I've used SI units which m/s is and mph is not.

I'd like to congratulate you on asking this question because one must always ask themselves whether the answer they've gotten is a reasonable one. Perhaps you were bothered for the same reason as I but then didn't take if to the next step and "What should my impression be of how high it should go?"

Last edited by Pmb; Jul 17th 2017 at 07:07 AM.   Jul 17th 2017, 10:40 AM   #6
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Thanks for your help it is much appreciated.

 Since you're setting v = 0 so you should get t = -v_0/a.
I found this way of doing the problem by looking at some websites, and it didn't make very much sense to me because plugging in a negative acceleration renders a negative result. So I think I understand what you're saying, but I'm still a little bit confused because I suppose I never fully understood the equation very well. I see how plugging v = 0 into v = at + V_0 and shifting it around will give t = -V_0 / a but what does the v and the V_0 represent? I'm not very clear on where the v = 0 comes from. Also, in my course I believe we have been usually using V_1 for the initial velocity and V_2 as the next and so on, so could you also please clarify what the V_0 represents? I assume it is the 2.249, but I'm just unclear on where the zero comes from.

Thanks again for your help!    Jul 17th 2017, 11:01 AM   #7
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 Originally Posted by jpompey Thanks for your help it is much appreciated. I found this way of doing the problem by looking at some websites, and it didn't make very much sense to me because plugging in a negative acceleration renders a negative result. So I think I understand what you're saying, but I'm still a little bit confused because I suppose I never fully understood the equation very well. I see how plugging v = 0 into v = at + V_0 and shifting it around will give t = -V_0 / a but what does the v and the V_0 represent? I'm not very clear on where the v = 0 comes from. Also, in my course I believe we have been usually using V_1 for the initial velocity and V_2 as the next and so on, so could you also please clarify what the V_0 represents? I assume it is the 2.249, but I'm just unclear on where the zero comes from. Thanks again for your help! The "0" subscript on a variable like that means "initial" i.e. its value at t = 0. In this case v(0) = v_0

To be precise: the velocity (a vector quantity) of a body as a function of time is written in two different ways, i.e. v = v(t). If the object is undergoing constant acceleration a (also a vector quantity) and the initial value of its velocity is v_0 then

v = v(t) = at + v_0

Let v be the vertical component of the velocity and v_0 the initial value of the horizontal component of velocity. Then

v = at + v_0

We want to find the time it takes for the body to reach its highest value. The vertical component of the velocity at that time t is zero. The horizontal component of the velocity is constant.

v = 0 = at + v_0

Solve for t

t = -9.8 m/s

My apologies for not recognizing that notation. Its uncommon but not unheard of.

Last edited by Pmb; Jul 17th 2017 at 11:03 AM.   Jul 17th 2017, 11:11 AM #8 Junior Member   Join Date: Jun 2017 Posts: 15 I live in Canada, we might use some slightly different signs than in the USA. Which would also explain the m/s Thanks again!   Jul 17th 2017, 11:14 AM   #9
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 Originally Posted by jpompey I live in Canada, we might use some slightly different signs than in the USA. Which would also explain the m/s Thanks again!
I can't imagine why anybody would use different signs since that's not a difference in notation but a fact of the motion.  Tags acceleration, displacement, time, velocity, vertical Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post toranc3 Kinematics and Dynamics 2 Jan 29th 2013 08:51 PM bradycat Kinematics and Dynamics 11 Feb 20th 2011 05:11 AM kristiandjess General Physics 0 Mar 29th 2009 11:55 AM jonbrutal Kinematics and Dynamics 1 Feb 2nd 2009 02:36 PM Kace_d Periodic and Circular Motion 1 Oct 6th 2008 10:40 AM