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Initial velocity in projectile motion
  • Thread starter Thread starter maverick99
  • Start date Start date Feb 9, 2005
  • Tags Tags Initial Initial velocity Motion Projectile Projectile motion Velocity
Click For Summary To find the initial velocity of a ball shot from a gun, the key equations involve time and distance. The time of flight can be calculated using t = sqrt(2h/g), where h is the height and g is the acceleration due to gravity (9.81 m/s² or 980 cm/s²). The initial horizontal velocity is then found using Vox = s/t, where s is the horizontal distance traveled. After substituting the measured values of s (284.5 cm) and h (76.5 cm) into these equations, the calculated initial velocity is 720 cm/sec. This process emphasizes the importance of consistent units throughout calculations. maverick99 Messages 74 Reaction score 0 I'm supposed to use g=9.81 m/sec and my measured values of s and h to find the initial velocity of a ball shot out of a gun i did in lab today. I was wondering what equation i use to find the initial velocity. Physics news on Phys.org
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dextercioby Science Advisor Insights Author Messages 13,403 Reaction score 4,156 I'm sure you meant g=9.8 ms^{-2}. This equation is the best: \vec{v}(t)=\vec{v}_{0}+\vec{g}t Daniel. maverick99 Messages 74 Reaction score 0 Could you explain what each of those letters represent? Integral Staff Emeritus Science Advisor Homework Helper Gold Member Messages 7,226 Reaction score 66 Perhaps one that relates the initial velocity to the the distance traveled and the time of flight? Seems like you need to examine the equations you have at your disposal. Maybe make and effort to look thorough your book and gather all of the motion equations together. Study the equations with emphasis on learning the meaning of each term. Here is the most basic equation, from which all can be derived. \ddot {x} = - g Where x increases up. Did that help? Perhaps not, as you have not provided even so basic of information as to what level of course this is. Please provide more information. Could you show us the "candidate" equations? What are you given? maverick99 Messages 74 Reaction score 0
Integral said: Perhaps one that relates the initial velocity to the the distance traveled and the time of flight? Seems like you need to examine the equations you have at your disposal. Maybe make and effort to look thorough your book and gather all of the motion equations together. Study the equations with emphasis on learning the meaning of each term. Here is the most basic equation, from which all can be derived. \ddot {x} = - g Where x increases up. Did that help? Perhaps not, as you have not provided even so basic of information as to what level of course this is. Please provide more information. Could you show us the "candidate" equations? What are you given?
Heres what the previous steps Since the initial velocity is only in the x-direction, v=vox, is vox =s/t and t can be determined from the y equation since the y displacemnt is equal to the height and is due solely to the acceleration of gravity: t=square root of (2h/g). Use g=9.81m/sec and your measured values of s and h to find the initial velocity of the ball. s= 284.5cm, h=76.5cm I'm also given h=1/2gt(squared) Integral Staff Emeritus Science Advisor Homework Helper Gold Member Messages 7,226 Reaction score 66 You are getting there. We are now even getting a glimmer of the experiment you did! Seems like you have the equation and the information requiered to compute the time.
t=square root of (2h/g).
Or in our Latex capable fourm t = \sqrt { \frac {2h} g (click on the equation to see what I typed to get that) Use that time to get your last result. Edit: Whoops I just recalled that you need V_0 Surely you have some equations relating V_0 to your other values? What is that equaition? Last edited: Feb 9, 2005 maverick99 Messages 74 Reaction score 0
Integral said: You are getting there. We are now even getting a glimmer of the experiment you did! Seems like you have the equation and the information requiered to compute the time. Or in our Latex capable fourm t = \sqrt { \frac {2h} g (click on the equation to see what I typed to get that) Use that time to get your last result. Edit: Whoops I just recalled that you need V_0 Surely you have some equations relating V_0 to your other values? What is that equaition?
I see 2 equations. Vox = s/t r = Vot + 1/2at(squared) maverick99 Messages 74 Reaction score 0 Here's something else too The initial velocity components are: Vox = Vo(Cos angle) and Voy = Vo(Sin angle) dextercioby Science Advisor Insights Author Messages 13,403 Reaction score 4,156 Yeah,then we could say you got the picture right. :smile: Daniel. xanthym Science Advisor Messages 410 Reaction score 0
maverick99 said: Here's something else too The initial velocity components are: Vox = Vo(Cos angle) and Voy = Vo(Sin angle)
However, you said before that the INITIAL velocity had no vertical component and was just horizontal. Thus, Voy=0. Try to focus on your original set of data and equations: (your words are quoted below)
Since the initial velocity is only in the x-direction, v=vox, is vox =s/t and t can be determined from the y equation since the y displacemnt is equal to the height and is due solely to the acceleration of gravity: t=square root of (2h/g). Use g=9.81m/sec and your measured values of s and h to find the initial velocity of the ball. s= 284.5cm, h=76.5cm I'm also given h=1/2gt(squared)
t = sqrt(2*h/g) Vox = s/t Do you see the technique?? maverick99 Messages 74 Reaction score 0
xanthym said: However, you said before that the INITIAL velocity had no vertical component and was just horizontal. Thus, Voy=0. Try to focus on your original set of data and equations: (your words are quoted below) t = sqrt(2*h/g) Vox = s/t Do you see the technique??
So is Vox = s/t the equation i need to find the initial velocity? xanthym Science Advisor Messages 410 Reaction score 0 That equation requires values for "s" and "t". You measured "s". How would you determine "t" in that equation?? maverick99 Messages 74 Reaction score 0
xanthym said: That equation requires values for "s" and "t". You measured "s". How would you determine "t" in that equation??
use the equation t=2*h/g? xanthym Science Advisor Messages 410 Reaction score 0
use the equation t=2*h/g?
CORRECT! Now substitute your measured values in the appropriate equations, and solve for Vox. Give it a try! (Show all your work.) ~ Last edited: Feb 9, 2005 maverick99 Messages 74 Reaction score 0
xanthym said: CORRECT! Now substitute your measured values in the appropriate equations, and solve for Vox. Give it a try! (Show all your work.) ~
Vox = s/t 284.5cm/3.95s i ended up getting 72, would the units be cm/sec? Last edited: Feb 9, 2005 xanthym Science Advisor Messages 410 Reaction score 0
maverick99 said: i ended up getting 72, would the units be cm/sec?
Check your UNITS. (Hint #1: Are ALL your calculation units in cm?) (Hint #2: g) maverick99 Messages 74 Reaction score 0 yah it's all in cm except for the gravity which is 9.81 I'm not sure how to put the gravity into it... xanthym Science Advisor Messages 410 Reaction score 0 The quickest method is to convert g to "cm/sec^2" from it current "m/sec^2": g = (9.8 m/sec^2) = (980 cm/sec^2) You need to repeat ALL calculations with everything the same EXCEPT now use g=(980) in the first equation. Please show all your work. ~ Last edited: Feb 9, 2005 maverick99 Messages 74 Reaction score 0 t = sqrt(2*h/g) Vox = s/t so, h=1/2gtsqred, so h=1/2(981cm/s)(3.95s)squared which is 7653 t=sqrt(2*7653/980cm/s) which is 3.95s Vox = s/t so, 284.5cm/3.95s which finally turns out to be 72 is this right?? xanthym Science Advisor Messages 410 Reaction score 0
maverick99 said: t = sqrt(2*h/g) Vox = s/t so, h=1/2gtsqred, so h=1/2(981cm/s)(3.95s)squared which is 7653 t=sqrt(2*7653/980cm/s) which is 3.95s Vox = s/t so, 284.5cm/3.95s which finally turns out to be 72 is this right??
NO. It is not correct this time. You originally performed the calculations correctly except that "g" had the wrong units. Let's return to your original calculation technique: THE EQUATIONS: Equation #1 -----> t = sqrt(2*h/g) Equation #2 -----> Vox = s/t YOUR MEASURED VALUES: s= 284.5cm, h=76.5cm STEP #1: Calculate "t" from Eq #1 using your MEASURED value of h=(76.5cm) and the known constant g=(980 cm/sec^2) STEP #2: Calculate "Vox" from Eq #2 using your MEASURED value of s=(284.5cm) and the value of "t" calculated from Step #1. You are very close to the correct answer. Take your time and focus on the steps above. ~~ Last edited: Feb 10, 2005 maverick99 Messages 74 Reaction score 0
xanthym said: NO. It is not correct this time. You originally performed the calculations correctly except that "g" had the wrong units. Let's return to your original calculation technique: THE EQUATIONS: Equation #1 -----> t = sqrt(2*h/g) Equation #2 -----> Vox = s/t YOUR MEASURED VALUES: s= 284.5cm, h=76.5cm STEP #1: Calculate "t" from Eq #1 using your MEASURED value of h=(76.5cm) and the known constant g=(980 cm/sec^2) STEP #2: Calculate "Vox" from Eq #2 using your MEASURED value of s=(284.5cm) and the value of "t" calculated from Step #1. You are very close to the correct answer. Take your time and focus on the steps above. ~~
k i did the steps and got 720, is that right? if so, 720 what? i don't know what the units would be xanthym Science Advisor Messages 410 Reaction score 0
maverick99 said: k i did the steps and got 720, is that right? if so, 720 what? i don't know what the units would be
CORRECT! Congratulations! The initial velocity is (720 cm/sec) since all your units were in "cm" and "sec". ~~ maverick99 Messages 74 Reaction score 0
xanthym said: CORRECT! Congratulations! The initial velocity is (720 cm/sec) since all your units were in "cm" and "sec". ~~
alright cool, thanks for the help!

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