Physics Help Forum Vertical displacement

 Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum

 Jul 16th 2017, 04:36 PM #1 Junior Member   Join Date: Jun 2017 Posts: 13 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: 13 thanks very much!
 Jul 17th 2017, 05:51 AM #4 Member     Join Date: Aug 2008 Posts: 80 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.

 Originally Posted by Pmb 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.

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: 13 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

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