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Net velocity of thrown objectlost 
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#1
Jan2206, 10:42 AM

P: 15

hey
I'm currently learning about velocities of objects thrown horizontally.The main point is that initial horizontal speed v[0] will remain constant while vertical velocity will accelerate with a = 10m / s^2. But when learning formulas there are few things that got me confused: 1)We can calculate net velocity of falling object moment before it hits the ground with: h > height v^2 = v[0]^2 + g^2*t^2 = v[0]^2 + 2*g*h We got g^2*t^2 = 2*g*h since t^2 = 2h / g Do we use this formula for net velocity also when an object is thrown in vertical direction with initial speed of v[0]? If that is the case then at least from my understanding this formula doesn't make much sense: *Firstly, t^2 = 2h / g is true only when object is dropped (and its initial velocity v[0] is 0 => v^2 = v[0]^2 + 2*g*h = 2*g*h) *If we toss objects vertically with initial speed of v[0],then its final net velocity should be : v = v[0] + g*t,and not v=sqrt( v[0]^2 + g^2*t^2 ) 2)And if formula for net velocity v[0]^2 + g^2*t^2 is indeed the same no matter if object is tossed with v[0] horizontally or vertically from same height, then it would suggest net velocity in both cases is the same. Which again doesn't make much sense to me! thank you for your help 


#2
Jan2206, 11:23 AM

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P: 39,564

Neither one is "net velocity". They are "vertical component" of the velocity vector. It is, in fact, true that the vertical component of velocity is independent of the horizontal component.



#3
Jan2206, 02:10 PM

P: 15

I'm sorry but your reply was way too scarce to help me with any of my questions



#4
Jan2206, 02:41 PM

HW Helper
P: 2,277

Net velocity of thrown objectlost
What is with this net velocity? You mean final speed (you're using scalars equations).
What type of movement you're describing? Projectile Motion? well it doesn't matter in any case most movements in a rectangular system can be solved by parametric equations, where the parameter is time (t). 


#5
Jan2206, 03:40 PM

P: 15

when object is thrown horizontally,final net velocity is vector object has moment before object hits the ground,and this vector is tangent to the path of object. When I'm talking about object being thrown vertically(meaning straight down) with speed V[0],I use word net velocity to describe vertical velocity. 


#6
Jan2206, 04:36 PM

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P: 41,475

The key to understanding this is: (1) Understand the kinematics of accelerated motion and how the equations are derived and interconnect with each other. (2) Realize that vertical and horizontal motions are independent, and that only the vertical is accelerated. 


#7
Jan2206, 05:53 PM

P: 15

Can you also reason with me in words(not formulas) why final speed of an object that falls distance h will be the same no matter in which direction object is thrown with initial speed v[0]? 


#8
Jan2306, 03:34 PM

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P: 41,475

Start with what we know (taking up to be positive): [tex]v = v_0 + a t = v_0 g t[/tex] Also, distance = average speed x time gives us: [tex]h = \frac{(v + v_0)}{2} t[/tex] (note the minus sign since it falls a distance h below the starting point) or: [tex]t = 2h/(v + v_0)[/tex] Plugging that into the first equation, we get: [tex]v = v_0 + a t = v_0 + 2g h/(v + v_0)[/tex] Multiply both sides by [itex](v + v_0)[/itex] and simplify to get: [tex]v^2 = v_0^2 + 2 g h[/tex] 


#9
Jan2306, 05:22 PM

P: 15

Off topic but I feel it's an important subject:
I'm not shure I understand why h is negative and you still get the same correct result. Is it because you also made acceleration vector g negative( [tex]v = v_0 + a t = v_0 g t[/tex] )? I know it's off topic, but when you decide to give some variables negative prefix, do you have to be consistent by also making certain other variables negative in order to derive correct result? Can you tell me some rules I can follow when doing that? And besides ,why is it important to mark h negative? So we know the direction of an object is downwards? Back to the subject at hand: But what if besides having initial vertical speed it would also have initial horizontal speed [tex]v_h[/tex]? Would formula then be [tex]v^2 = v_h^2 + v_0^2 + 2 g h [/tex] ? I imagine not since from it we can't derive [tex]v^2 = v_0^2 + 2 g h[/tex] where [tex]v_0^2[/tex] would represent initial speed (initial vertical and initial horizontal combined). Yup,I'm lost :( 


#10
Jan2306, 07:40 PM

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P: 41,475

[tex]v^2 = v_{0,h}^2 + v_{0,v}^2 + 2 g h[/tex] So, we've proven, as I think you understand, that: [tex]v_v^2 = v_{0,v}^2 + 2 g h[/tex] And you know that: [tex]v_h^2 = v_{0,h}^2[/tex] Add them up to find the "net" final velocity after falling a distance h: [tex]v^2 = v_h^2 + v_v^2 = v_{0,h}^2 + v_{0,v}^2 + 2 g h[/tex] Which gives us: [tex]v^2 = v_0^2 + 2 g h[/tex] where [itex]v_0^2[/itex] is the total initial speed squared, regardless of direction (since we added the components to get the total). Got it? 


#11
Jan2306, 07:44 PM

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P: 41,475

[tex]v^2 = v_{0,h}^2 + v_{0,v}^2 + 2 g h[/tex] So, we've proven, as I think you understand, that: [tex]v_v^2 = v_{0,v}^2 + 2 g h[/tex] And you know that: [tex]v_h^2 = v_{0,h}^2[/tex] Add them up to find the "net" final velocity after falling a distance h: [tex]v^2 = v_h^2 + v_v^2 = v_{0,h}^2 + v_{0,v}^2 + 2 g h[/tex] Which gives us: [tex]v^2 = v_0^2 + 2 g h[/tex] where [itex]v_0^2[/itex] is the total ("net") initial speed squared, regardless of direction (since we added the components to get the total). Got it? 


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